Unit 1: Dynamics
Are You Ready?, pages 4–5
1. The displacement of the student who takes the bus is the vector sum of each displacement
along the journey of the bus. The displacement of the student who rides her bicycle is the single
vector from the starting position to school. The final displacement of each student is the vector
from his or her starting position to the school. Displacement depends only on initial and final
position.
2. The baseball player has more acceleration because he starts from rest and speeds up. The jet
has no acceleration because it moves at constant velocity.
3. Answers may vary. Sample answer: A car moving east but slowing down has its velocity
vector [E] in the opposite direction to its acceleration vector [W].
4. Answers may vary. Sample answer:
(a) Flicking water from your hands is an example of Newton’s first law: your hands stop
suddenly but the water continues to move forward.
(b) An effective way to get ketchup out of a bottle is to shake the bottle upside down. By
Newton’s first law, the ketchup has a tendency to keep moving when the bottle stops. There is
little friction between the bottle and the ketchup, and Newton’s second law states that the
ketchup will not slow down with the bottle.
(c) When you hit the bottom of a ketchup bottle you apply a force to the bottle and, through the
bottle, to the ketchup inside. Since you are holding the bottle tightly, there is no net force on the
bottle. By Newton’s second law, the bottle will not accelerate. Whether the ketchup moves is
determined by the forces holding the ketchup in the bottle and the relative masses of the bottle
and ketchup. If the tap on the bottle is strong enough to break the surface tension between the
bottle and the ketchup, the ketchup may feel a small net force and move with respect to the
bottle.
5. Answers may vary. Sample answer: The car experiences a forward applied force from the man
and the man feels a backward normal force from the car. These forces are an action–reaction
pair. Friction in the rolling mechanism of the car makes it more difficult for the man to get the
car moving, while static friction between the man and the ground allows the man to exert a
strong applied force to the car.
6. Answers may vary. Sample answer:
(a) The forces the players exert on each other are an action–reaction pair. By Newton’s third law,
the forces are equal in magnitude and opposite in direction.
(b) By Newton’s second law, the smaller player has a greater acceleration because he has a
smaller mass.
(c) The force a player experiences in a collision may break a bone or cause serious bruising or
internal injury. By Newton’s first law, the parts of his body not hit tend to stay at rest while the
struck parts will accelerate. Protective equipment shields the body from direct contact and allows
the force to act on the player as a whole.
7. (a) Initially, static friction keeps the heavy box from moving. The static friction is equal in
magnitude and opposite in direction to the applied force.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-1
!
!
(b) Four forces act on the box when it is moving: the force of gravity, Fg , normal force, FN , an
!
!
applied force, Fa , and kinetic friction, Ff .
(c) Graphs may vary. Sample graph:
While the box is at rest, the static friction is equal in magnitude to the applied force. When static
friction reaches its breaking point, the box begins to move. While the box is moving, kinetic
friction has a constant value determined by the weight of the box and the nature of the surfaces
involved.
9. (a) System diagrams may vary. Students should draw a side view of a puck sliding along an
ice surface and coming to rest. A sample diagram is shown.
(b)
10. Answers may vary. Sample answer: I could use a range finder set in velocity mode and
similar plastic containers filled with various masses. For each container, I would record its speed
as it fell from close to the ceiling to the floor and then examine the v–t graph. To test Aristotle’s
first proposition, I would check if the speed of the containers jumped immediately to a value that
then remained constant. To test Aristotle’s second proposition, I would see if the containers
reached a terminal velocity and, if so, whether that velocity was different for differing masses. If
any one of the containers showed reasonably constant acceleration for any part of its fall, both
propositions would be disproved.
11. By Newton’s second law, acceleration is inversely proportional to mass for a fixed force. Plot
(a) shows this inverse relationship.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-2
12. Given: ΔtTP = 0.75 h; ΔdTP = 90 km; ΔdPO = 220 km
Required: arrival time in Ottawa based on ΔtPO
Analysis: The storm is moving at reasonably constant velocity. Calculate the speed of the storm
!d
from its trip from Toronto to Peterborough, v = TP . Then determine the time to Ottawa
!tTP
using v =
!d PO
!d PO
; ΔtPO =
.
v
!tPO
Solution: v =
=
!dTP
!tTP
90 km
0.75 h
v = 120 km/h
!tPO =
=
!d PO
v
220 km
120 km /h
= 1.833 h "
60 min
1 h
!tPO = 100 min
Statement: The storm arrives in Ottawa at 6:10 p.m., 100 min after passing through
Peterborough at 4:30 p.m.
!
13. (a) Given: v = 41.0 km/h [N]; !t = 15 min
Required: distance travelled by the boat, !d
Analysis: Use the given speed and time interval to determine the distance using v =
Δd = vΔt.
Solution:
!d = v!t
= 41.0
!d
;
!t
km
1 h
" 15 min "
h
60 min
!d = 1.0 " 101 km
Statement: The boat will travel 1.0 × 101 km in 15 min.
(b) The net force acting on the boat is 0 N because the boat is travelling at a constant speed.
(c) The total of all the frictional forces acting on the boat is equal in magnitude and opposite in
direction to the applied
force of !the water on the boat.
!
14. (a) Given: FA = 55 N [W]; FB = 82 N [E]
!
Required: net force, Fnet
!
!
!
Analysis: The net force is the vector sum of the individual forces, Fnet = FA + FB . Use east as
positive in the calculations.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-3
!
!
!
Solution: Fnet = FA + FB
= !55 N + 82 N
! = +27 N
Fnet = 27 N [E]
Statement: The
! net force is 27 N [E].
(b) Given: Fnet = +27 N; m = 12 kg
!
Required: acceleration ofr the mass, a
r
r r F
Analysis: Fnet = ma; a = net
m
r
r F
Solution: a = net
m
+27 N
=
12 kg
= +2.2 m/s 2
r
a = 2.2 m/s 2 [E]
Statement: The acceleration of the mass is 2.2 m/s2 [E].
15. (a) Given: m = 14 kg
Required: the magnitude of all the forces acting on the mass!
Analysis: Two forces act on the mass: the force of gravity, Fg , and the normal force of the
!
desk, FN . The force of gravity acts down and is the weight of the mass, Fg = mg . The normal
!
!
force balances the force of gravity, FN = ! Fg , because the mass is at rest.
Solution: Fg = mg
= (14 kg)(9.8 m/s 2 )
Fg = 140 N
FN = Fg
FN = 140 N
Statement: The magnitude of the force of gravity is 140 N, and the magnitude of the normal
force is 140 N. The net force
is 0 N.
!
(b) Given: m = 3.2 kg; Fa = 4.5 N [forward] ; g = 9.8 m/s2
Required: the magnitude of all the forces acting on the mass
!
Analysis: There are four forces acting on the mass: the force of gravity, Fg ; the normal force of
!
!
!
the floor, FN ; the applied force, Fa ; and the force of friction, Ff . The force of gravity acts down
and is the weight of the mass, Fg = mg . The normal force balances the force of gravity,
!
!
FN = ! Fg , because the mass is moving horizontally. The mass is moving at a constant velocity,
!
!
so the force of friction balances the applied force, Ff = ! Fa .
Solution: Fg = mg
= (3.2 kg)(9.8 m/s 2 )
Fg = 31 N
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-4
FN = Fg
FN = 31 N
Ff = Fa
Ff = 4.5 N
Statement: The magnitude of the force of gravity is 31 N, of the normal force is 31 N, of the
applied force is 4.5 N, and of the force of friction is 4.5 N. The net force is 0 N.
!
!
(c) Given: m = 4.7 kg; Fa = 8.6 N [forward]; a = 1.1 m/s 2 [forward] ; g = 9.8 m/s2
Required: the magnitude of all the forces acting on the mass
!
Analysis: There are four forces acting on the mass: the force of gravity, Fg ; the normal force of
!
!
!
the floor, FN ; the applied force, Fa ; and the force of friction, Ff . The mass is accelerating
horizontally. The force of gravity and the normal force balance, but there is a net force in the
forward direction. Use Newton’s second law, Fnet = ma , to determine the magnitude of the net
force and then the force of friction. Use forward as positive.
Solution: Fg = mg
FN = Fg
= (4.7 kg)(9.8 m/s 2 )
Fg = 46 N
Fnet = ma
= (4.7 kg)(1.1 m/s 2 )
FN = 46 N
!
! !
Fnet = Fa + Ff
Ff = Fnet ! Fa
Fnet = 5.17 N (one extra digit carried)
= 5.17 N ! 8.6 N
Ff = 3.4 N
Statement: The magnitude of the force of gravity is 46 N, of the normal force is 46 N, of the
applied force is 8.6 N, and of the force of friction is 3.4 N. The net force is 5.2 N [forward].
16. (a) Given: m = 15 kg; g = 9.8 m/s2
Required: FN
Analysis: The scale reads the magnitude of the normal force exerted by the pan of the scale. The
pan is at rest, so the normal force balances the weight of the mass.
Solution: Fg = mg
FN = Fg
2
FN = 150 N
= (15 kg)(9.8 m/s )
Fg = 150 N
Statement: The scale reads
150 N.
!
(b) Given: m = 15 kg; Fa = 22 N [down] ; Fg = 150 N
Required: FN
Analysis: The scale reads the magnitude of the normal force exerted by the pan of the scale. The
pan is at rest, so the net force on the pan is zero. Determine the normal force using Fnet = 0 N.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-5
Solution: Fnet = FN ! Fg ! Fa
FN = Fnet + Fg + Fa
= 0 N + 150 N + 22 N
FN = 170 N
Statement: The scale reads 170 N.
(c) The magnitude of the normal force acting on the mass is the reading on the scale. By
Newton’s third law, the force of the scale on the mass (normal force) and the force of the mass
on the scale are equal in magnitude.
17. Analysis: The triangles are right-angled triangles. Use the basic trigonometric ratios to solve
for the unknown lengths.
Solution: The hypotenuse is given (17 m) and side a is adjacent to the given angle (52°).
a
cos 52° =
17 m
a = (17 m)(cos 52°)
a = 10.5 m
The hypotenuse is given (17 m) and side b is opposite the given angle (52°).
b
sin 52° =
17 m
b = (17 m)(sin 52°)
b = 13.4 m
Side c is the hypotenuse, and the side adjacent to the given angle is known (25 m; 37°).
25 m
cos 37° =
c
25 m
c=
cos 37°
c = 31.3 m
Side d is opposite the given angle (37°), and the adjacent side is known (25 m).
d
tan 37° =
25 m
d = (25 m)(tan 37°)
d = 18.8 m
Statement: The unknown lengths are a = 10.5 m, b = 13.4 m, c = 31.3 m, and d = 18.8 m.
18. Given: a = 12 cm; b = 9.0 cm; ! C = 21°
Required: the length of c and the measures of angles x and y.
Analysis: The triangle is not a right triangle, so use the cosine law, c 2 = a 2 + b2 ! 2abcosC , to
sin A sin B sinC
=
=
determine the length of c. Use the sine law,
, to determine angle x. Use the
b
c
a
sum of the internal angles of a triangle to determine the measure of angle y.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-6
Solution: c 2 = a 2 + b2 ! 2abcosC
= (12 cm)2 + (9.0 cm)2 ! 2(12 cm)(9.0 cm)(cos 21°)
c 2 = 23.35 cm 2
c = 4.832 cm (two extra digits carried)
c = 4.8 cm
sin x sinC
=
a
c
sin x
sin 21°
=
9.0 cm 4.832 cm
(9.0 cm )(sin 21°)
sin x =
4.832 cm
sin x = 0.6675 (two extra digits carried)
x = sin !1 (0.6675)
= 41.9° (one extra digit carried)
x = 42°
y + x + C = 180°
y + 41.9° + 21° = 180°
y = 117°
Statement: The length of side c is 4.8 cm. The measure of angle x is 42° and the measure of
angle y is 117°.
Copyright © 2012 Nelson Education Ltd.
Unit 1: Dynamics
U1-7
Section 1.1: Motion and Motion Graphs
Tutorial 1 Practice, page 10
1. (a) Given: !d1 = 1.2 km; !t1 = 24 min; !d2 = 1.2 km; !t2 = 24 min
Required: vav
Analysis: Calculate the total distance travelled, !d = !d1 + !d2 , and the total time
taken, !t = !t1 + !t2 . Then, use the equation vav =
!t1 = 24 min "
1h
!d
to calculate the average speed.
!t
60 min
!t1 = 0.40 h
!t2 = 0.40 h
Solution: !d = !d1 + !d2
= 1.2 km + 1.2 km
!d = 2.4 km
!t = !t1 + !t2
= 0.40 h + 0.40 h
!t = 0.80 h
!d
vav =
!t
2.4 km
=
0.80 h
vav = 3.0 km/h
Statement: The
r average speed for the entire route is 3.0 km/h.
(b) Given: !d = 1.2 km [E]; !t = 24min
!
Required: vav
!
!d
!
Analysis: The average velocity is the ratio of the displacement and the time taken, vav =
.
!t
1h
!t = 24 min "
60 min
!t = 0.40 h
!
!d
!
Solution: vav =
!t
1.2 km [E]
=
0.40 h
!
vav = 3.0 km/h [E]
Statement: The average velocity from the house to the farthest position from the house is
3.0 km/h [E].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-1
r
r
(c) Given: !d1 = 1.2 km [E]; !t1 = 0.40 h; !d2 = 1.2 km [W]; !t2 = 0.40 h
!
Required: vav
Analysis: Since the two displacements are equal in magnitude and opposite in direction, the total
displacement during the walk is 0.0 km. So, the average velocity is 0.0 km/h.
Statement: The average velocity is 0.0 km/h.
(d) Answers may vary. Sample answer: The average velocities in parts (b) and (c) are different.
Average velocity depends on displacement, which is a vector quantity. Since the displacements
are not the same, the average velocities are not the same.
!
2. Given: v = 27 m/s [forward]; !t = 0.32 s
!
Required: !d
!
r r r
!d
!
Analysis: Rearrange vav =
to solve for !d: !d = vav !t.
!t
r r
Solution: !d = vav !t
= (27 m/s [forward])(0.32 s)
r
!d = 8.6 m [forward]
Statement: The bus moves 8.6 m [forward] before the driver reacts.
3. Given: !d = 200 laps, at 4.02 km/lap; !t = 6.69 h
Required: vav in km/h
Analysis: Calculate the distance covered (in kilometres) and then the average speed, vav =
Solution: !d = 200 laps "
4.02 km
1 lap
!d = 804 km
!d
.
!t
!d
!t
804 km
=
6.69 h
vav = 1.20 " 102 km/h
vav =
Statement: The average speed of a driver who completes 200 laps in 6.69 h is 1.20 ! 102 km/h .
4. (a) Given: !d1 = 140 m; !t1 = 55 s; !d2 = 45 m; !t2 = 21 s
r
Required: vav
Analysis: Calculate the total distance walked, !d = !d1 + !d2 , and the time
taken, !t = !t1 + !t2 . Then, calculate the average speed using vav =
Solution: !d = !d1 + !d2
= 140 m + 45 m
!d = 185 m
!t = !t1 + !t2
= 55 s + 21 s
!t = 86 s
Statement: The student’s average speed is 2.4 m/s.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
!d
.
!t
!d
!t
185 m
=
76 s
vav = 2.4 m/s
vav =
1.1-2
!
!
(b) Given: !d1 = 140 m [E]; !t1 = 55 s; !d2 = 45 m [W]; !t2 = 21 s
r
Required: vav
!
!
!
Analysis: Calculate the total displacement, !d = !d1 + !d2 . Then, calculate the average velocity
!
!d
!
using vav =
. Use east as positive.
!t
!
!
!
!
!d
!
Solution: !d = !d1 + !d2
vav =
!t
= 140 m + ("45 m)
95 m [E]
=
= +95 m
76 s
!
!
!d = 95 m [E]
vav = 1.2 m/s [E]
Statement: The student’s average velocity is 1.2 m/s [E].
!
!
5. (a) Given: !d1 = 62 km [S]; !d2 = 78 m [N]; vav = 55 km/h
!
Required: vav
!d
Analysis: Use the total distance, !d = !d1 + !d2 , and average speed, vav =
, to determine the
!t
!
!
!
time taken for the trip, !t . Then, determine the total displacement, !d = !d1 + !d2 . Then
!
!d
!
calculate the average velocity using vav =
. Use north as positive.
!t
!d
Solution: !d = !d1 + !d2
vav =
!t
= 62 km + 78 km
!d
!t =
= 140 km
vav
!d = 140 km
140 km
=
55 km /h
!t = 2.545 h (two extra digits carried)
!
!d
!
vav =
!t
= ("62 km) + (+78 km)
16 km [N]
=
= +16 km
2.545 h
!
!
!d = 16 m [N]
vav = 6.3 km/h [N]
Statement: The truck’s average velocity is 6.3 km/h [N].
(b) Answers may vary. Sample answer: The truck turned around at one point during its trip. At
the end, it was only 16 km away from where it started. As a result, the magnitude of the average
velocity is quite low compared to the average speed.
!
!
!
!d = !d1 + !d2
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-3
Tutorial 2 Practice, page 13
1. (a) Velocity is slope on a position–time graph. Graph (c) has increasing slope, showing that
the velocity is increasing.
(b) Graph (b) has decreasing slope, showing that the velocity is decreasing.
2. Answers may vary. Sample answers:
(a) Initially, the velocity is large and constant in the east direction. Then, it is lower and constant
in the east direction. Calculate the slope of each section of the position–time graph using data
from the graph.
t (s)
0.0
0.2
0.8
!
0.0
20
30
d (m [E])
!
!
___
r
30 " 20
!d 20 " 0.0
!d
v (m/h [E])
=
= 100
=
= 17
!t 0.2 " 0.0
!t 0.8 " 0.2
Use these velocity data to draw a velocity–time graph.
(b) Initially the velocity is large and constant in a west direction, then it is zero and finally it is
low and constant in an east direction. Calculate the slope of each section of the position–time
graph using data from the graph.
0.0
0.5
1.0
2.0
t (h)
!
0.0
15
15
0.0
d (m [W])
!
!
!
___
!
15 " 15
!d 15 " 0.0
!d
!d 0.0 " 15
v (m/h [W])
=
= 30
=
=0
=
= "15
!t 0.5 " 0.0
!t 1.0 " 0.5
!t 2.0 " 1.0
Use these velocity data to draw a velocity–time graph.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-4
(c) Initially the velocity is low and constant in a north direction, and then it is large and constant
in a south direction. Calculate the slope of each section of the displacement–time graph using
data from the graph.
0.0
4.0
8.0
t (h)
r
50
100
!50
d (m [S])
!
!
___
r
!d ("50) " 50
!d 100 " ("50)
v (m/h [S])
=
= "25
=
= 37.5
4.0 " 0.0
8.0 " 4.0
!t
!t
Use these velocity data to draw a velocity–time graph.
Tutorial 3 Practice, page 15
1. Answers may vary. Sample answer:
(a) The velocity–time graph is a straight line showing that the car’s velocity is changing at a
constant rate. So, the car is moving with constant acceleration.
(b) The car starts from rest at t = 0 s. It moves north with increasing speed and constant
acceleration. The car is moving at 12 m/s after 6.0 s.
(c) Given: velocity–time graph
!
Required: a
Analysis: Read the coordinates of two points on the graph. Use these points to calculate the
!
! !v
slope of the line, a =
. As in the graph, use north as positive.
!t
Solution: Two clear points are (0.0 s, 0 m/s) and (6.0 s, 12 m/s).
!
! !v
a=
!t
12 m/s " 0 m/s
=
6.0 s " 0.0 s
12 m/s
=
6.0 s
= +2.0 m/s 2
!
a = 2.0 m/s 2 [N]
Statement: The acceleration of the car is 2.0 m/s2 [N].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-5
2. Answers may vary. Sample answer:
(a) Given: velocity–time graph
!
Required: aav
Analysis: Read the coordinates of the initial and final points on the graph. Use these points to
!
!v
!
. As in the graph, use forward as positive.
calculate the slope of the secant, aav =
!t
Solution: The points are (0 s, 0 m/s) and (6 s, 35 m/s).
!
!v
!
aav =
!t
35 m/s " 0 m/s
=
6 s"0 s
= +6 m/s 2
!
aav = 6 m/s 2 [forward]
Statement: The average acceleration of the car for the entire trip is 6 m/s2 [forward].
(b) Given: velocity–time graph
!
Required: a at t = 3 s and at t = 5 s.
Analysis: The instantaneous acceleration at a given time is the slope of the tangent to the
velocity–time graph at that time. Sketch tangent lines at the required times, read the coordinates
of two points on each tangent line, and then calculate the slopes.
Solution:
For the tangent at 3 s, two points are
For the tangent at 5 s, two points are
(1.5 s, 0 m/s) and (6 s, 25 m/s).
(2.5 s, 0 m/s) and (6 s, 34 m/s).
r
r
r !v
r !v
a=
a=
!t
!t
25 m/s " 0 m/s
34 m/s " 0 m/s
=
=
6 s " 1.5 s
6 s " 2.5 s
25 m/s
34 m/s
=
=
4.5 s
3.5 s
r
r
2
a = 6 m/s [forward]
a = 10 m/s 2 [forward]
Statement: The instantaneous acceleration is 6 m/s2 [forward] at 3 s and 10 m/s2 [forward] at 5 s.
(c) The slope of the velocity–time graph increases gradually, so the acceleration–time graph
should be increasing. Use the values of instantaneous acceleration from part (b) to draw the graph.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-6
3. Answers may vary. Sample answer:
(a) Given: velocity–time graph
!
Required: aav
Analysis: Read the coordinates of initial and final points on the graph. Use these points to
!v
. As in the graph, use forward as positive.
calculate the slope of the secant, aav =
!t
Solution: The points are (0 s, 50 m/s) and (7 s, 0 m/s).
!v
aav =
!t
0 m/s " 50 m/s
=
7 s"0 s
"50 m/s
=
7s
a = "7 m/s 2
Statement: The average acceleration of the car for the entire trip is 7 m/s2 [backward].
(b) Given: velocity–time graph
!
Required: a at t = 2 s, t = 4 s, and t = 6 s
Analysis: The instantaneous acceleration at a given time is the slope of the tangent to the
velocity–time graph at that time. Sketch tangent lines at the required times, read the coordinates
of two points on each tangent line, and then calculate the slopes.
Solution:
For the tangent at 2 s, two points are
For the tangent at 4 s, two points are
(1 s, 50 m/s) and (3 s, 42 m/s).
(3 s, 42 m/s) and (5 s, 26 m/s).
!
!
! !v
! !v
a=
a=
!t
!t
42 m/s " 50 m/s
26 m/s " 42 m/s
=
=
3 s "1 s
5s"3s
"8 m/s
"16 m/s
=
=
2s
2s
!
!
2
a = "4 m/s
a = "8 m/s 2
For the tangent at 6 s, two points are (5 s, 26 m/s) and (7 s, 2 m/s).
!
! !v
a=
!t
2 m/s " 26 m/s
=
7 s"5 s
"24 m/s
=
2s
!
a = "12 m/s 2
Statement: The instantaneous acceleration is 4 m/s2 [backward] at 2 s, 8 m/s2 [backward] at 4 s,
and 12 m/s2 [backward] at 6 s.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-7
(c) The slope of the velocity–time graph decreases gradually, so the acceleration–time graph
should be decreasing, possibly linearly. Use the values of instantaneous acceleration from
part (b) to draw the graph.
Section 1.1 Questions, page 16
1. (a) Given: !d1 = 22 m; !d2 = 11 m
Required: total distance, !d
Analysis: !d = !d1 + !d2
Solution: !d = !d1 + !d2
= 22 m + 11 m
!d = 33 m
Statement: The total distance travelled by the cardinal is 33 m.
(b) Given: !t1 = 2.9 s; !t2 = 1.5 s
Required: vav
!d
. To determine the total time add the partial times, !t = !t1 + !t2 .
!t
Solution: !t = !t1 + !t2
Analysis: vav =
= 2.9 s + 1.5 s
!t = 4.4 s
!d
!t
33 m
=
4.4 s
vav = 7.5 m/s
Statement: The average speed of the cardinal is 7.5 m/s.
vav =
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-8
!
!
(c) Given: !d1 = 22 m [E]; !d2 = 11 m [N]
!
Required: vav
!
!d
!
Analysis: The average velocity is the ratio of the total displacement to the total time, vav =
.
!t
To determine the total displacement, calculate the vector sum of the partial displacements,
!
!
!
!d = !d1 + !d2 .
Solution: Determine the magnitude of the displacement.
! 2
! 2
!2
!d = !d1 + !d2
= (22 m)2 + (11 m)2
2
! = 605 m
!d = 24.60 m (two extra digits carried)
Determine the angle ! .
!
"d2
tan ! = !
"d1
11 m
22 m
tan ! = 0.50
=
! = tan #1 (0.50)
! = 27°
Determine the average velocity.
!
!d
!
vav =
!t
24.60 m [E 27° N]
=
4.4 s
!
vav = 5.6 m/s [E 27° N]
Statement: The cardinal’s average velocity is 5.6 m/s [E 27° N] .
2. (a) Given: vav = 15.0 km/h
Required: vav in m/s
Analysis: Convert units using 1 km = 1000 m, and 1 h = 3600 s.
km
Solution: vav = 15.0
h
km 1000 m
1h
= 15.0
!
!
3600 s
h
1 km
= 4.167 m/s (one extra digit carried)
vav = 4.17 m/s
Statement: The average speed is 4.17 m/s.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-9
(b) Given: !d = 2.90 km; vav = 4.167 m/s
Required: !t
!d
!d
Analysis: Use the relation vav =
.
; !t =
!t
vav
Solution: !t =
=
!d
vav
2.90 km 1000 m
"
4.167 m/s 1 km
!t = 696 s
Statement: It takes 696 s to complete one lap.
3. (a) Given: !d = 16 m; !t = 2.1 s
Required: average speed, vav
!d
!t
!d
Solution: vav =
!t
16 m
=
2.1 s
vav = 7.6 m/s
Statement: The average speed of the skater is 7.6 m/s.
(b) Given: diameter, D = 16 m
Required: time for one lap, !t
Analysis: vav =
Analysis: Calculate the circumference of the pond, !d = " D . Use vav =
time, !t , for one lap.
Solution: !d = " D
= " (16 m)
!d = 50.26 m (two extra digits carried)
!d
to determine the
!t
!d
!t
!d
!t =
vav
vav =
50.26 m
7.619 m/s
!t = 6.6 s
Statement: It takes the skater 6.6 s to go around the edge of the pond.
=
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-10
4. (a) Given: !d = 450 km = 4.5 " 104 m; !t = 45 min
Required: vav , in m/s
Analysis: vav =
!t = 45 min "
!d
;
!t
60 s
1 min
3
!t = 2.7 " 10 s
!d
Solution: vav =
!t
4.5 " 104 m
=
2.7 " 103 s
vav = 170 m/s
Statement: The airplane’s average speed is 170 m/s.
!
(b) Given: !d = 450 km [E 15° N]
!
Required: vav , in m/s
Analysis: The airplane travels in one direction for the whole trip, so the average velocity is a
vector equal in magnitude to the average speed.
Solution: The airplane’s average velocity is 170 m/s [E 15° N] .
r
r
5. Given: vi = 0 m/s; vf = 96 km/h [W]; !t = 4.1 s
!
Required: a
Analysis: Since the rocket starts from rest, its change in velocity is equal to its final velocity:
!
! !v
! !
.
!v = vf . Its acceleration is the ratio of its change in velocity to the time interval taken: a =
!t
1h
1 min
! 96 km 1000 m
!
!
!
vf =
h
60 s
60 min
1 km
!
vf = 26.67 m/s [W] (one extra digit carried)
!
! !v
Solution: a =
!t
26.67 m/s [W]
=
4.1 s
!
a = 6.5 m/s 2 [W]
Statement: The rocket accelerates at 6.5 m/s 2 [W] .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-11
!
!
6. Given: aav = 1.37 ! 103 m/s 2 [W]; "t = 3.12 ! 10#2 s; vf = 0 m/s
r
Required: velocity of the ball before hitting the wall, vi
Analysis: The ball slows down when it hits the wall. Since the acceleration points west, the
!
!v
!
initial velocity must point east. Use aav =
to determine the change in velocity and the initial
!t
velocity.
!
! !v
Solution: a =
!t
! !
!v = aav !t
= (1.37 " 103 m/s 2 [W])(3.12 " 10#2 s)
!
!v = 42.744 m/s [W] (two extra digits carried)
r r r
!v = vf " vi
r
r r
vi = vf " !v
= 0 m/s " 42.744 m/s [W]
r
vi = 42.7 m/s [E]
Statement: The ball hits the wall with velocity 42.7 m/s [E] .
!
!
7. Given: vi = 0 m/s; vf = 9.3 m/s [forward]; !t = 3.9 s
!
Required: average acceleration of the runner, aav
!
!v
!
Analysis: aav =
!t
!
!v
!
Solution: aav =
!t
9.3 m/s [forward] " 0 m/s
=
3.9 s
!
2
aav = 2.4 m/s [forward]
Statement: The runner’s average acceleration is 2.4 m/s2 [forward].
8. Answers may vary slightly based on student reading of the graph data. Sample answer:
(a) Given: position–time graph
!
Required: vav for the entire trip
Analysis: Read the coordinates of initial and final points on the graph. Use these points to
!d
calculate the slope of the secant, vav =
. As in the graph, use east as positive.
!t
Solution: The points are (0 s, 0 m) and (14 s, 400 m).
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-12
!d
!t
400 m " 0 m
=
14 s " 0 s
400 m
=
14 s
vav = 29 m/s
Statement: The average velocity of the car for the entire trip is 29 m/s2 [E].
(b) Given: position–time graph
!
Required: vav from t = 4 s to t = 14 s
Analysis: Use the graph to determine the position at t = 4 s and at t = 14 s. Use these points to
!d
calculate the slope of the secant, vav =
. As in the graph, use east as positive.
!t
Solution: The points are (4 s, 30 m) and (14 s, 400 m).
!d
vav =
!t
400 m " 30 m
=
14 s " 4 s
370 m
=
10 s
vav = 37 m/s
Statement: The average velocity of the car for the last 10 s of the trip is 37 m/s [E]. This is
higher than for the trip as a whole because the car is accelerating—the car was moving faster on
average toward the end of the trip.
(c) Given: position–time graph
!
Required: v at t = 4.0 s, t = 8.0 s, and at t = 12.0 s
Analysis: The instantaneous velocity at a given time is the slope of the tangent to the position–
time graph at that time. Sketch in tangent lines at the required times, read the coordinates of two
points on each tangent line and then calculate the slopes.
Solution:
For the tangent at 4.0 s, two points are
For the tangent at 8.0 s, two points are
(2.0 s, 0.0 m/s) and (6.0 s, 68 m/s).
(6.0 s, 70 m/s) and (10 s, 200 m/s).
!d
!d
v=
v=
!t
!t
68 m " 0 m
200 m " 70 m
=
=
6.0 s " 2.0 s
10.0 s " 6.0 s
68 m
130 m
=
=
4.0 s
4.0 s
v = 17 m/s
v = 33 m/s
vav =
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-13
For the tangent at 12 s, two points are (10.0 s, 200 m/s) and (14.0 s, 400 m/s).
!d
v=
!t
400 m " 200 m
=
14.0 s " 10.0 s
200 m
=
4.0 s
v = 50 m/s
Statement: The instantaneous velocities are approximately 17 m/s [E] at 4.0 s, 33 m/s [E] at
8.0 s, and 50 m/s [E] at 12.0 s.
(d) The slope of the position–time graph gradually increases throughout the trip showing that the
velocity is increasing to the east. The velocity–time graph should show this gradual increase; it
may be linear. Draw a velocity–time graph using the values of instantaneous velocity. Draw a
smooth line for a reasonable trend.
9. Answers may vary slightly based on student reading of the graph data. Sample answer:
(a) Given: velocity–time graph
!
Required: aav
Analysis: Read the coordinates of the initial and final points on the graph. Use these points to
!v
calculate the slope of the secant, aav =
. As in the graph, use forward as positive.
!t
Solution: The points are (0 s, 50 m/s) and (10 s, 0 m/s).
!v
aav =
!t
0 m/s " 50 m/s
=
10 s " 0 s
"50 m/s
=
10 s
aav = "5 m/s 2
Statement: The average acceleration of the car for the entire trip is 5 m/s2 [backward].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-14
(b) Given: velocity–time graph
!
Required: a at t = 3 s, t = 6 s and at t = 9 s
Analysis: The instantaneous acceleration at a given time is the slope of the tangent to the
velocity–time graph at that time. Sketch in tangent lines at the required times, read the
coordinates of two points on each tangent line and then calculate the slopes.
Solution:
For the tangent at 3 s, two points are
For the tangent at 6 s, two points are
(2 s, 50 m/s) and (4 s, 44 m/s).
(5 s, 36 m/s) and (7 s, 24 m/s).
!
!
! !v
! !v
a=
a=
!t
!t
44 m/s " 50 m/s
24 m/s " 36 m/s
=
=
4 s"2 s
7 s"5 s
"6 m/s
"12 m/s
=
=
2s
2s
!
!
2
a = "3 m/s
a = "6 m/s 2
For the tangent at 9 s, two points are (8 s, 18 m/s) and (10 s, 0 m/s).
!v
a=
!t
0 m/s " 18 m/s
=
10 s " 8 s
"18 m/s
=
2s
a = "9 m/s 2
Statement: The instantaneous accelerations are 3 m/s2 [backward] at 3 s, 6 m/s2 [backward] at
6 s, and 9 m/s2 [backward] at 9 s.
(c) The slope of the velocity–time graph decreases gradually, so the acceleration–time graph
should be decreasing, possibly linearly. Use the values of instantaneous acceleration from
part (b) to draw the graph.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.1-15
Section 1.2: Equations of Motion
Tutorial 1 Practice, page 19
!
!
!
1. (a) Given: vi = 15.0 m/s [forward]; vf = 0 m/s; a = 5.0 m/s 2 [backward]
!
Required: !d
! !
! !
1!
! v ! vi
Analysis: Use a = f
to determine the braking time, !t . Then use !d = vi !t + a!t 2 to
"t
2
determine the braking distance. Choose forward to be the positive direction.
Solution:
v ! vi
1
!d = vi !t + a!t 2
a= f
2
"t
1
v ! vi
= (15.0 m/s)(3.0 s) + ("5.0 m/ s 2 )(3.0 s )2
"t = f
2
a
= 45.0 m " 22.5 m
0 m/s ! 15.0 m/s
=
2
!5.0 m/s
!d = 22 m
"t = 3.0 s
Statement: The motorcycle’s braking distance is 22 m [forward].
(b) Answers may vary. Sample answer: In Sample Problem 2 and part (a), I calculated the
braking distance for a motorcyclist slowing down at 5.0 m/s2. When the initial speed is high, it
takes much longer and a much greater distance to stop, so a speeding vehicle has more difficulty
stopping safely.
!
!
2. Given: vi = 0 m/s; !d = 120 m [N]; !t = 15 s
!
Required: a
! !
1!
Analysis: Use !d = vi !t + a!t 2 to determine the acceleration.
2
1
!d = vi !t + a!t 2
2
1
!d = (0 m/s)!t + a!t 2
2
1
!d = a!t 2
2
2!d
a= 2
!t
Choose north to be the positive direction.
2!d
Solution: a = 2
!t
2(120 m)
=
(15 s)2
a = 1.1 m/s 2
Statement: The runner accelerates at 1.1 m/s2 [N].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-1
!
!
!
3. Given: v1 = 22 m/s [E]; !t1 = 12 s; a = 1.2 m/s 2 [W]; vf = 0 m/s
!
Required: total displacement, !d
Analysis: Determine the displacement for the first part of the trip using
!
! !d1
v1 =
!t1
!d1 = v1!t1
Determine the braking time using
! !
! vf ! vi
a=
"t
vf ! vi
"t =
a
! !
1!
Then, use !d2 = vi !t + a!t 2 to determine the displacement for the second part of the trip.
2
!
!
!
Finally, the total displacement is !d = !d1 + !d2 . Use east as the positive direction.
Solution: !d1 = v1!t1
= (22 m/s)(12 s)
!d1 = 264 m (one extra digit carried)
vf " vi
a
0 m/s " 22 m/s
=
"1.2 m/s 2
!t = 18.33 s (two extra digits carried)
!t =
1
!d2 = v1!t + a!t 2
2
1
= (22 m/s)(18.33 s) + ("1.2 m/ s 2 )(18.33 s) 2
2
!d2 = 202 m (one extra digit carried)
!d = !d1 + !d2
= 264 m + 202 m
!d = 470 m
Statement: The total displacement of the bus is 4.7 ! 102 m .
!
!
4. (a) Given: vi = 0 m/s; vf = 9.6 m/s [W]; !t = 4.2 s
!
Required: a
! !
! vf ! vi
Analysis: Calculate the acceleration using a =
. Use west as positive.
"t
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-2
vf ! vi
"t
9.6 m/s ! 0 m/s
=
4.2 s
= 2.286 m/s 2 (two extra digits carried)
Solution: a =
a = 2.3 m/s 2
Statement: The runner’s acceleration is 2.3 m/s2 [W].
!
!
(b) Given: vi = 0 m/s; !t = 4.2 s; a = 2.286 m/s 2 [W]
!
Required: !d
! !
1!
Analysis: Calculate the displacement using !d = vi !t + a!t 2 .
2
1
Solution: !d = vi !t + a!t 2
2
1
= (0 m/s)(4.2 s) + (2.286 m/ s 2 )(4.2 s) 2
2
= 20.16 m (two extra digits carried)
!d = 20 m
Statement: The displacement of the runner is 2.0 × 101 m [W] while accelerating.
!
!
!
(c) Given: v = 9.6 m/s [W]; !d = 100.0 m [W]; !d1 = 20.16 m [W]; !t1 = 4.2 s
Required: total time, !t
!
!
!
!
Analysis: Use !d = !d1 + !d2 to calculate the displacement, !d2 , for the second part of the
race.
!d = !d1 + !d2
!d2 = !d " !d1
!
!d
!
Then use the constant velocity formula, vav =
, to determine the time, !t2 , for the second part
!t
of the race.
!d
v2 = 2
!t2
!t2 =
!d2
v2
Finally, the total time taken is the sum of the two times, !t = !t1 + !t2 .
Solution: !d2 = !d " !d1
= 100.0 m " 20.16 m
!d2 = 79.84 m (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
!t2 =
!d2
v2
79.84 m
9.6 m/s
!t2 = 8.317 s (two extra digits carried)
=
Chapter 1: Kinematics
1.2-3
!t = !t1 + !t2
= 4.2 s + 8.317 s
!t = 13 s
Statement: The total time taken is 13 s.
!
!
5. (a) Given: player 1: d1i = 0 m [right]; v1i = 0 m/s; a = 2.4 m/s 2 [right] ;
!
!
player 2: d2i = 42 m [right]; v2 = 5.4 m/s [left]
Required: time until the players meet, !t
Analysis: Set up an equation for each player that relates position and time. The first player
! !
1!
moves at constant acceleration, !d = vi !t + a!t 2 .
2
The second
player
moves
at
a
constant
velocity:
!
! !d
v=
!t
!d
v2 = 2
!t
!d2 = v2 !t
Compare these equations to solve for either position or time. Use right as positive.
"b ± b2 " 4ac
Solve the quadratic equation using the quadratic formula: !t =
.
2a
player 2:
Solution: player 1:
1
!d = v1i !t + a!t 2
2
1
!d1 = (0 m/s)!t + (2.4 m/s 2 )!t 2
2
2
d1f " d1i = (1.2 m/s )!t 2
!d2 = v2 !t
d2f " d2i = ("5.4 m/s)!t
d2f " 42 m = ("5.4 m/s)!t
d2f = 42 m " (5.4 m/s)!t
d1f = (1.2 m/s 2 )!t 2
The final positions of the two players are equal.
d1f = d2f
(1.2 m/s 2 )!t 2 = 42 m " (5.4 m/s)!t
(1.2 m/s 2 )!t 2 + (5.4 m/s)!t " 42 m = 0
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-4
This is a quadratic equation in !t .
!t =
=
"b ± b2 " 4ac
2a
"5.4 m/s ± (–5.4 m/s)2 " 4(1.2 m/s 2 )(–42 m)
2(1.2 m/s 2 )
!t = 4.1 s or " 8.8 s
The second solution is not possible, because the time taken to run is positive.
Statement: The players collide after 4.1 s.
(b) Given: equation for player 1: d1 = (1.2 m/s 2 )!t 2 ; equation for player 2: !d2 = v2 !t ;
!t = 4.1 s
!
!
Required: !d1; !d2
Analysis: Use the time taken by the players and the player’s equations of motion to determine
their displacements.
Solution: player 1:
player 2:
2
2
!d2 = v2 !t
!d1 = (1.2 m/s )!t
= (1.2 m/s 2 )(4.1 s)2
= ("5.4 m/s)(4.1 s)
!d2 = "22 m
!d1 = 20 m
Statement: Player 1 ran 20 m [right] and player 2 ran 22 m [left].
!
!
(c) Given: v1i = 0 m/s; a = 2.4 m/s 2 ; !t = 4.1 s
Required: v1f
! !
! vf ! vi
Analysis: Use a =
to solve for the final speed of player 1.
"t
v1 f ! v1i = a"t
v1 f = v1i + a"t
Solution: v1 f = v1i + a!t
= 0 m/s + (2.4 m/s 2 )(4.1 s)
v1 f = 9.8 m/s
Statement: Player 1 is moving at 9.8 m/s when he collides with player 2.
!
!
!
6. (a) Given: v1 = 110 m/s [forward]; v2 = 0 m/s; a = 6.2 m/s 2 [backward]
Required: minimum stopping time, !t
! !
! vf ! vi
Analysis: a =
; Use forward as positive.
"t
v ! vi
"t = f
a
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-5
vf " vi
a
0 m/s " 110 m/s
=
"6.2 m/s 2
= 17.74 s (two extra digits carried)
Solution: !t =
!t = 18 s
Statement: The airplane needs a minimum of 18 s to stop.
!
!
(b) Given: vi = 110 m/s; !t = 17.74 s; a = 6.2 m/s 2 [backward]
Required: minimum stopping distance, !d
1
Analysis: Use !d = vi !t + a!t 2 to determine the distance travelled while stopping.
2
1
Solution: !d = vi !t + a!t 2
2
1
= (110 m/s)(17.74 s) + ("6.2 m/ s 2 )(17.74 s) 2
2
2
!d = 9.8 # 10 m
Statement: The minimum safe length of the runway is 9.8 ! 102 m .
(c) Answers may vary. Sample answer: The runway should be much longer than the minimum
length because the airplane may land partway along the runway. If the weather is bad, the usual
braking force needed to slow the plane to a stop may be insufficient and more time and distance
may be needed to stop the plane. Also, it is uncomfortable for the passengers to slow a plane
down using the maximum allowed braking force.
Tutorial 2 Practice, page 20
!
!
!
1. Given: vi = 22 m/s [up]; vf = 0 m/s; a = 9.8 m/s 2 [down]
Required: maximum height, !d
Analysis: Use vf2 = vi2 + 2a!d to calculate !d .
vf2 = vi2 + 2a!d
vf2 " vi2
2a
Use up as positive and down as negative.
vf2 " vi2
Solution: !d =
2a
(0 m/s) 2 " (22 m/s) 2
=
2("9.8 m/ s 2 )
!d =
!d = 25 m
Statement: The maximum height of the apple is 25 m.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-6
!
!
! 1
2. Given: vi = 0 m/s; !d = 10.0 m [down]; a = (9.8 m/s 2 [down])
6
Required: time for an object to fall, !t
1
Analysis: Calculate the acceleration and then use !d = vi !t + a!t 2 to determine !t .
2
"b ± b2 " 4ac
Solve the quadratic equation using the quadratic formula: !t =
.
2a
Use up as positive.
!9.8 m/s 2
Solution: a =
6
a = !1.633 m/s 2 (two extra digits carried)
1
!d = vi !t + a!t 2
2
1
"10.0 m = (0 m/s)!t + ("1.633 m/s 2 )!t 2
2
2
2
("0.8165 m/s )!t + (0 m/s)!t + 10.0 m = 0
!t =
=
"b ± b2 " 4ac
2a
(
)
0 m/s ± (0 m/s)2 " 4 "0.8165 m/s 2 (10.0 m)
"1.633 m/s
2
!t = 3.5 s
Statement: It would take 3.5 s for an object to fall 10.0 m if g were one-sixth the value of
Earth’s g.
!
!
!
3. (a) Given: vi = 12 m/s [down]; !d = 45 m [down]; a = 9.8 m/s 2 [down]
Required: time to fall, !t
1
Analysis: Use up as positive. Use !d = vi !t + a!t 2 to determine !t .
2
1
a!t 2 + vi !t " !d = 0
2
Use the quadratic formula to determine Δt.
#1 &
"vi ± vi2 " 4 % a ( ("!d)
$2 '
!t =
a
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-7
#1 &
"vi ± vi2 " 4 % a ( ("!d)
$2 '
Solution: !t =
a
"("12 m/s) ± ("12 m/s)2 " 4("4.9 m/s 2 )(45 m)
"9.8 m/s 2
= 2.044 s or " 4.493 s (two extra digits carried)
=
!t = 2.0 s
Statement: The ball takes 2.0 s to land.
!
!
!
(b) Given: vi = 12 m/s [down]; !d = 45 m [down]; a = 9.8 m/s 2 [down]; !t = 2.044 s
Required: vf
Analysis: Use a =
"v +v %
vf ! vi
or !d = $ i f ' !t to determine vf . Both solutions are provided.
"t
# 2 &
vf ! vi
"t
vf ! vi = a"t
a=
vf = vi + a"t
"v +v %
!d = $ i f ' !t
# 2 &
2!d
!t
2!d
vf =
( vi
!t
vi + vf =
Solution: vf = vi + a!t
= ("12 m/s) + ("9.8 m/s 2 )(2.044 s)
= "12 m/s " 20.03 m/s
vf = "32 m/s
2!d
" vi
!t
2("45 m)
=
" ("12 m/s)
2.044 s
= "44.03 m/s + 12 m/s
vf = "32 m/s
vf =
Statement: The ball is moving at 32 m/s when it lands.
(c) Answers may vary. Sample answer: If the ball were thrown up at 12 m/s, it would rise to a
maximum height and then fall. It would be moving at 12 m/s down as it passed its initial
position. It would then continue down as in the question. The ball would be moving at the same
speed as above, 32 m/s.
!
!
4. (a) Given: !d = 32 m [down]; !t = 1.5 s; a = 9.8 m/s 2 [down]
!
Required: velocity at 32 m, vi
Analysis: The displacement and time taken are known. Calculate average velocity using
! " v! + v! %
!d = $ i f ' !t . This gives the sum of the velocities.
# 2 &
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-8
"v +v %
!d = $ i f ' !t
# 2 &
vi + vf =
2!d
!t
The acceleration is known, and the difference in the (squares of the) velocities can be determined
using vf2 = vi2 + 2a!d . The initial velocity of the ball can then be determined.
vf2 = vi2 + 2a!d
vf2 " vi2 = 2a!d
Although it is not usual, for convenience, use down as the positive direction.
2!d
Solution: vi + vf =
!t
2(32 m)
=
1.5 s
vi + vf = 42.67 m/s (two extra digits carried) (Equation 1)
vf2 ! vi2 = 2a"d
(
)
= 2 9.8 m/s 2 (32 m)
vf2 ! vi2 = 627.2 (m/s)2 (two extra digits carried) (Equation 2)
The left side of Equation 2 can be used to determine the difference of the speeds.
vf2 ! vi2 = 627.2 (m/s)2
(vf ! vi )(vf + vi ) = 627.2 (m/s)2
(vf ! vi )(42.67 m/s) = 627.2 (m/s)2
vf ! vi = 14.70 m/s (two extra digits carried) (Equation 3)
Finally, use equations 1 and 3 to determine vi .
vf ! vi = 14.70 m/s
(Equation 3)
vi ! vf = !14.70 m/s
vi + vf = 42.67 m/s
(Equation 1)
2vi = !14.70 m/s + 42.67 m/s
vi = 14 m/s
Statement: The ball is moving 14 m/s [downward] when the timing starts at 32 m.
(b) Solutions may vary. Sample answer:
!
!
!
Given: vi = 0 m/s; vf = 14 m/s [downward]; !d2 = 32 m; a = 9.8 m/s 2 [downward]
!
Required: total displacement, !d
Analysis: Calculate the displacement for the first part of the trip using:
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-9
vf2 = vi2 + 2a!d
!d1 =
vf2 " vi2
2a
Then calculate the total displacement using !d = !d1 + !d2 . Continue to use down as the
positive direction.
v 2 " vi2
Solution: !d1 = f
!d = !d1 + !d2
2a
= 10 m + 32 m
(14 m/s)2 " (0 m/s)2
=
!d = 42 m
2(9.8 m/ s 2 )
!d1 = 10 m
Statement: The ball’s displacement is 42 m [downward].
!
!
5. (a) Given: !d = 14 m [up]; !t = 1.1 s; a = 9.8 m/s 2 [down]
!
Required: vi
Analysis: Use up as positive. Calculate the initial velocity using
1
!d = vi !t + a!t 2
2
1
!d " a!t 2
2
vi =
!t
1
!d " a!t 2
2
Solution: vi =
!t
1
14 m " ("9.8 m/s 2 )(1.1 s)2
2
=
1.1 s
14 m + 5.93 m
=
1.1 s
= 18.12 m/s (two extra digits carried)
vi = 18 m/s
Statement: The initial velocity of the keys was 18 m/s [up].
(b) Solutions may vary. Sample answer:
!
!
!
Given: !d = 14 m [up]; a = 9.8 m/s 2 [downward]; vi = 18.12 m/s
!
Required: vf
Analysis: Use vf2 = vi2 + 2a!d to determine vf . (Any of the equations containing vf could be
used.)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-10
Solution: vf2 = vi2 + 2a!d
= (18.12 m/s)2 + 2("9.8 m/s 2 )(14 m)
= 328.33 (m/s)2 " 274.4 (m/s)2
= 53.93 (m/s)2
vf = 7.3 m/s
Statement: The velocity of the keys when they were caught was 7.3 m/s [up].
Section 1.2 Questions, page 21
!
!
2
1. (a) Given: vi = 0 m/s; !t = 5.2 s; a = 4.1 m/s [forward]
!
Required: !d
! !
1!
Analysis: Calculate the displacement using !d = vi !t + a!t 2 . Use forward as the positive
2
direction.
1
Solution: !d = vi !t + a!t 2
2
1
= (0 m/s)(5.2 s) + (4.1 m/ s 2 )(5.2 s) 2
2
!d = 55 m
Statement: The racehorse’s displacement is 55 m [forward].
!
!
(b) Given: vi = 0 m/s; !t = 5.2 s; a = 4.1 m/s 2 [forward]
!
Required: vf
Analysis: Use a =
vf ! vi
"t
vf = vi + a"t
vf ! vi
to determine vf :
"t
a=
(Any of the equations containing vf could be used.)
Solution: vf = vi + a!t
= 0 m/s + (4.1 m/s 2 )(5.2 s)
vf = 21 m/s
Statement: The horse’s final velocity is 21 m/s [forward].
!
!
!
2. (a) Given: vi = 7.72 ! 106 m/s [E]; vf = 2.46 ! 106 m/s [E]; "d = 0.478 m [E]
!
Required: a
Analysis: Use east as the positive direction.
Use vf2 = vi2 + 2a!d to calculate the acceleration:
vf2 = vi2 + 2a!d
a=
vf2 " vi2
2!d
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-11
vf2 ! vi2
2"d
(2.46 # 106 m/s)2 ! (7.72 # 106 m/s)2
=
2(0.478 m)
Solution: a =
!5.355 # 1013 m 2 /s 2
0.956 m
a = !5.60 # 1013 m/s 2
Statement: The electron’s acceleration is 5.60 ! 1013 m/s 2 [W].
!
!
!
(b) Given: vi = 7.72 ! 106 m/s [E]; vf = 2.46 ! 106 m/s [E]; a = 5.60 ! 1013 m/s 2 [W]
Required: !t
v ! vi
v " vi
Analysis: Use a = f
to determine !t : !t = f
"t
a
(Any of the equations containing !t could be used.)
v " vi
Solution: !t = f
a
(2.46 # 106 m/s) " (7.72 # 106 m/s)
=
"5.60 # 1013 m/s 2
"5.26 # 106 m/s
=
"5.60 # 1013 m/s 2
=
!t = 9.39 # 10"8 s
Statement: The acceleration occurs over a time interval of 9.39 ! 10"8 s .
3. Solutions may vary. Sample answer:
!
!
(a) Given: cruiser: d1i = 0 m [forward]; v1i = 0 m/s; a = 3.0 m/s 2 [forward] ;
!
!
car: d2i = 0 m [forward]; v2 = 62 km/h [forward]
Required: time until the cruiser catches up with the car, !t
Analysis: The vehicles start from the same position. The displacement and time when the two
vehicles meet are not known. Set up an equation for each vehicle that relates position and time.
! !
1!
The cruiser moves at constant acceleration, !d = vi !t + a!t 2 , and the car at constant velocity,
2
!
! !d
. Compare these equations to solve for time and, later, displacement. Use forward as
v=
!t
positive.
1h
1 min
! 62 km 1000 m
v2 =
!
!
!
1h
60 s
1 km
60 min
!
v2 = 17.22 m/s (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-12
Solution: cruiser:
1
!d1 = vi !t + a!t 2
2
1
= (0 m/s)!t + (3.0 m/s 2 )!t 2
2
2
!d1 = (1.5 m/s )!t 2
The displacements of the two vehicles are equal.
!d1 = !d2
car:
!d2
!t
!d2 = v2 !t
v2 =
!d2 = (17.22 m/s)!t
(1.5 m/s 2 )!t 2 = (17.22 m/s)!t
(1.5 m/s 2 )!t 2 = (17.22 m/s) !t
17.22 m/s
1.5 m/s 2
= 11.48 s (two extra digits carried)
!t =
!t = 11 s
Statement: The cruiser catches up 11 s after starting.
!
(b) Given: car: v2 = 17.22 m/s [forward]; !t = 11.48 s
Required: common displacement of the vehicles, !d
Analysis: Substitute the value of !t in the displacement equation of either the cruiser or the car.
Use the equation of the car.
Solution: !d2 = (17.22 m/s)!t
= (17.22 m/s)(11.48 s)
!d2 = 2.0 " 102 m
Statement: The cruiser catches up to the car 2.0 ! 102 m [forward] from the cruiser’s initial
position.
!
(c) Given: cruiser: v1i = 0 m/s; a = 3.0 m/s 2 [forward]; !t = 11.48 s
Required: v1f
! !
! vf ! vi
Analysis: Use a =
to solve for the final speed of the cruiser.
"t
v !v
a = 1f 1i
"t
v1f ! v1i = a"t
v1f = v1i + a"t
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-13
Solution: v1f = v1i + a!t
= (0 m/s) + (3.0 m/s 2 )(11.48 s)
= 34 m/s
34 m
1 km
60 s
60 min
=
"
"
"
s
1000 m 1 min
1h
v1f = 120 km/h
Statement: The cruiser is moving at 120 km/h when it catches up with the car. This is an unsafe
speed in a school zone.
r
r
(d) Given: cruiser: v1i = 0 m/s; v1f = 72 km/h [forward]; a = 3.0 m/s 2 [forward] ; then at constant
!
speed; car: v2 = 17.22 m/s [forward]
Required: time until the cruiser catches up with the car, !t
Analysis: Build a new equation for the displacement of the cruiser and the car.
72 km 1000 m
1h
1 min
v1f =
!
!
!
h
60 s
1 km
60 min
v1f = 20 m/s
Determine how long and how far the cruiser accelerates to reach the speed of 20 m/s. Use
vf = vi + a!t and vf2 = vi2 + 2a!d . Then write an equation for the cruiser’s displacement while
continuing at constant speed. Determine where the car is when the cruiser’s motion changes, and
then write a new equation for the car’s displacement. The rest of the solution will be similar to
part (a).
!
Solution: The cruiser’s final speed is vf = 20 m/s [forward]. It reaches 20 m/s when
vf = vi + a!t
vf " vi
a
20 m/s " 0 m/s
=
3.0 m/s 2
!t = 6.67 s
The cruiser’s displacement is now
vf2 = vi2 + 2a!d
!t =
vf2 " vi2
!d1 =
2a
(20 m/s)2 " (0 m/s)2
=
2(3.0 m/s 2 )
=
400 m 2 / s 2
6.0 m/ s 2
d1 = 66.7 m (one extra digit carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-14
At 6.67 s, the car’s displacement is
!d2 = v2 !t
!d2 = (17.22 m/s)(6.67 s)
d2 = 114.9 m (two extra digits carried)
Treat the motion as if it started at ti = 6.67 s , after which both vehicles move at constant speed.
The later positions of the vehicles are
cruiser:
car:
!d1 = v1!t
!d2 = v2 !t
d1 " 66.7 m = (20 m/s)!t
d2 " 114.9 m = (17.22 m/s)!t
d1 = 66.7 m + (20 m/s)!t
d2 = 114.9 m + (17.22 m/s)!t
The cruiser catches up to the car when the positions are equal.
d1 = d2
66.7 m + (20 m/s)!t = 114.9 m + (17.22 m/s)!t
(20 m/s)!t " (17.22 m/s)!t = 114.9 m " 66.7 m
48.2 m
2.78 m/s
!t = 17.34 s (two extra digits carried)
!t =
d1 = 66.7 m + (20 m/s)!t
= 66.7 m + (20 m/s)(17.34 s)
d1 = 413.5 m (two extra digits carried)
The total time elapsed from the start is
!t = 6.67 s + 17.34 s
!t = 24 s
Statement: Under the second scenario, it takes 24 s for the cruiser to catch the car, by which
time the cruiser and the car have travelled more than 400 m. They might be on the other side of
the school zone and it would be too late to protect the children. Neither of these scenarios is very
reasonable.
4. Answers may vary. Sample answer: I would stand at the top of the cliff with a small stone. My
partner would be at the bottom of the cliff with a stopwatch. I would signal with my hand the
instant I dropped the stone. My partner would time the stone’s fall. Since I would now know the
time interval taken, the acceleration due to gravity, and the initial velocity (0 m/s), I would use
! !
1!
!d = vi !t + a!t 2 to determine the displacement. The magnitude of the displacement would be
2
our estimate of the height of the cliff. The assumption I would have to make is that the height of
the cliff is being measured from sea level.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-15
5. Solutions may vary. Sample answer:
!
(a) Given: !t = 2.4 s; a = 9.8 m/s 2 [down]
!
Required: vi
Analysis: Since the ball is caught at the same height from which it is thrown, its total
! " v! + v! %
displacement is 0 m. From !d = $ i f ' !t , the final velocity is the opposite of the initial
# 2 &
! !
!
! v ! vi
velocity. Use this fact and a = f
to calculate vi .
"t
! !
! v ! vi
a= f
"t
!
!2
vi
!
a=
"t
!
a"t
!
vi = !
2
Use up as the positive direction.
!
a"t
!
Solution: vi = !
2
(!9.8 m/s 2 )(2.4 s)
=!
2
= 11.76 m/s (two extra digits carried)
!
vi = 12 m/s [up]
Statement: The ball’s initial velocity is 12 m/s [up].
!
!
(b) Given: !t = 2.4 s; a = 9.8 m/s 2 [down]; vi = 11.76 m/s [up]
Required: maximum height, !d
Analysis: The flight of the ball is symmetric going up and coming down, so the time to reach
maximum height is one-half of the total time interval, 2.4 s: !t = 1.2 s . The speed of the ball at
maximum height is 0 m/s. Use any of the formulas that contain !d to calculate the height. This
"v +v %
solution uses !d = $ i f ' !t .
# 2 &
"v +v %
Solution: !d = $ i f ' !t
# 2 &
(11.76 m/s + 0 m/s)(1.2 s)
2
!d = 7.1 m
Statement: The maximum height of the ball is 7.1 m.
=
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-16
6. Solutions may vary. Sample answer:
!
!
!
(a) Given: vi = 18 m/s [up]; !d = 32 m [down]; a = 9.8 m/s 2 [down]
Required: time to reach the ground, !t
1
Analysis: Use up as positive. Use !d = vi !t + a!t 2 to calculate !t :
2
1
a!t 2 + vi !t " !d = 0
2
Use the quadratic formula to determine Δt.
#1 &
"vi ± vi2 " 4 % a ( ("!d)
$2 '
!t =
a
1
"vi ± vi2 " 4( a)("!d)
2
Solution: !t =
a
"18 m/s ± (18 m/s)2 " 4("4.9 m/s 2 )(32 m)
=
–9.8 m/s 2
= 4.984 s (or "1.310 s) (two extra digits carried)
!t = 5.0 s
Statement: The ball takes 5.0 s to hit the ground.
!
!
!
(b) Given: vi = 18 m/s [up]; !d = 32 m [down]; a = 9.8 m/s 2 [down]; !t = 4.984 s
Required: vf
vf ! vi
to calculate vf = vi + a!t .
"t
Solution: vf = vi + a!t
= (18 m/s) + ("9.8 m/s 2 )(4.984 s)
= 18 m/s " 48.80 m/s
vf = "31 m/s
Analysis: Use a =
Statement: The velocity of the ball when it hits the ground is 3.1 × 101 m/s [down].
!
!
!
(c) Given: vi = 18 m/s [up]; vf = 0 m/s ; a = 9.8 m/s 2 [down]
Required: maximum height, dmax
vf2 " vi2
Analysis: Use v = v + 2a!d to solve for !d =
.
2a
Then !d = dmax " di to solve for dmax .
2
f
2
i
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-17
vf2 " vi2
2a
(0 m/s)2 " (18 m/s)2
=
2("9.8 m/s 2 )
Solution: !d =
=
324 m 2 / s 2
19.6 m/ s 2
!d = 16.53 m (two extra digits carried)
dmax = di + !d
= 32 m + 16.53 m
dmax = 49 m
Statement: The ball reaches a maximum height of 49 m.
(d) The flight of the ball is not symmetric. The time and distance travelled from my hand to the
maximum are smaller than the time and distance travelled from the maximum to the ground. So,
half the total time does not correspond to the time for either part of the whole flight.
7. Solutions may vary. Sample answer:
!
!
(a) Given: vi = 0 m/s; !t = 5.0 s; a = 39.2 m/s 2 [up]
!
Required: vf
vf ! vi
to determine vf = vi + a!t .
"t
Use up as the positive direction.
Solution: vf = vi + a!t
= 0 m/s + (39.2 m/s 2 )(5.0 s)
= 196 m/s (one extra digit carried)
vf = 2.0 " 102 m/s
Analysis: Use a =
Statement: The rocket is moving at 2.0 ! 102 m/s [up] when the engines stop.
!
!
!
(b) Given: vi = 196 m/s [up]; vf = 0 m/s; a = 9.8 m/s 2 [down]
Required: maximum height of the rocket, df
1
Analysis: Determine the rocket’s height when the engine shuts off using !d = vi !t + a!t 2 .
2
2
2
Then determine the displacement of the rocket while it slows down using vf = vi + 2a!d .
vf2 = vi2 + 2a!d
vf2 " vi2
2a
Together, these will give us the maximum height. Continue to use up as positive.
!d =
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-18
1
Solution: When the engine has finished firing, the rocket has risen !d = vi !t + a!t 2 .
2
1
!d = vi !t + a!t 2
2
1
= (0 m/s)(5.0 s) + (39.2 m/ s 2 )(5.0 s) 2
2
!d = 490 m
From when the engines stops to maximum height, the rocket goes up.
vf2 " vi2
!d =
2a
(0 m/s) 2 " (196 m/s) 2
=
2("9.8 m/ s 2 )
!d = 1960 m
The maximum height is
df = di + !d
= 490 m + 1960 m
= 2450 m
df = 2.4 " 103 m
Statement: The maximum height of the rocket is 2.4 ! 103 m .
!
!
!
(c) Given: vi = 0 m/s ; di = 2450 m [up] from part (b); a = 9.8 m/s 2 [down]
Required: time to reach the ground, !t , and total time of the flight, ttot
Analysis: Use vf = vi + a!t to calculate the time taken from the engine stopping to maximum:
!t =
vf " vi
1
. Then, use !d = vi !t + a!t 2 to determine the time to fall to the ground:
a
2
#1 &
"vi ± vi2 " 4 % a ( ("!d)
$2 '
!t =
a
Lastly, add the times needed for the three parts of the flight to determine the total time.
v " vi
Solution: !t = f
a
0 m/s " 196 m/s
=
"9.8 m/s 2
!t = 20 s
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-19
For the rocket to fall to the ground from its maximum height:
#1 &
"vi ± vi2 " 4 % a ( ("!d)
$2 '
!t =
a
0 ± 02 " 4("4.9)(2450)
=
"9.8
= ±22.36 s (two extra digits carried)
ttot = 5.0 s + 20 s + 22.36 s
ttot = 47 s
Statement: The time for the rocket to fall from rest at the maximum is 22 s. The total time from
being at rest initially is 47 s.
8. Solutions may vary. Sample answer:
Table 2
Acceleration
Reaction
Speed
Braking
(m/s2)
time (s)
(km/h)
distance (m)
(i)
9.5
0.80
60.0
28
(ii)
9.5
0.80
120.0
85
(iii)
9.5
2.0
60.0
48
!
!
(a) (i) Given: v1 = 60.0 km/h [forward]; !t1 = 0.80 s; a = 9.5 m/s 2 [backward]
Required: braking distance, d
!d
Analysis: The car moves at constant speed during the reaction time. Use vav =
to determine
!t
the distance covered during this time: !d = v!t .
The car then slows to a stop. Determine the distance covered during this time using:
vf2 = vi2 + 2a!d
vf2 " vi2
2a
Then, determine the total of these distances travelled. Use forward as the positive direction.
1h
1 min
! 60.0 km 1000 m
v1 =
!
!
!
1h
60 s
1 km
60 min
!
v1 = 16.67 m/s [forward] (two extra digits carried)
!d =
Solution: !d1 = v1!t
= (16.67 m/s ) (0.80 s)
!d1 = 13.34 m (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-20
!d2 =
=
vf2 " vi2
2a
(0 m/s)2 " (16.67 m/s)2
2("9.5 m/ s 2 )
!d = !d1 + !d2
= 13.34 m + 14.63 m
!d = 28 m
!d2 = 14.63 m (two extra digits carried)
Statement: The braking distance is 28 m.
r
(ii) Given: v2 = 120.0 km/h [forward] = 33.33 m/s [forward]; !t2 = 0.80 s;
!
a = 9.5 m/s 2 [backward]
Required: braking distance, d
Analysis: Repeat the steps of part (i).
vf2 " vi2
!d4 =
Solution: !d3 = v2 !t2
2a
= ( 33.33 m/s ) (0.80 s)
(0 )2 " (33.33 m/s)2
=
!d3 = 26.66 m (two extra digits carried)
2("9.5 m/ s 2 )
!d4 = 58.47 m (two extra digits carried)
d = !d3 + !d4
= 26.66 m + 58.47 m
d = 85 m
Statement: The braking distance is 85 m.
r
r
(iii) Given: v3 = 60.0 km/h [forward] = 16.67 m/s [forward]; !t3 = 2.0 s; a = 9.5 m/s 2 [backward]
Required: braking distance, d
Analysis: Repeat the steps of part (i).
Solution: !d5 = v3!t3
= (16.67 m/s )(2.0 s)
!d5 = 33.34 m (two extra digits carried)
v " vi2
!d6 =
2a
(0 m/s)2 " (16.67 m/s)2
=
2("9.5 m/ s 2 )
2
f
d = !d5 + !d6
= 33.34 m + 14.63 m
d = 48 m
!d6 = 14.63 m (two extra digits carried)
Statement: The braking distance is 48 m.
(b) If a driver uses a cellphone while driving, it is likely that the driver’s reaction time and
braking distance will increase. The table shows that at 60 km/h, increasing the reaction time from
0.80 s to 2.0 s causes the braking distance to increase from 28 m to 48 m (almost double).
Similarly, the driving speed affects the braking distance. For a reaction time of 0.80 s, increasing
the speed from 60 km/h to 120 km/h results in the braking distance increasing from 28 m to 85 m
(about triple).
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.2-21
Section 1.3: Displacement in Two Dimensions
Tutorial 1 Practice, page 25
!
!
1. Given: !d1 = 1.2 km [S]; !d2 = 3.1 km [E 53° N]
!
!
Required: !dT ; the angle for !dT , !
!
!
!
Analysis: !dT = !d1 + !d2 . Decide on a scale and then draw each vector to scale on a coordinate
axis. Draw the total displacement vector.
Solution: An appropriate scale is 1 cm : 0.3 km. Calculate the lengths of the arrows for the
displacement vectors:
r
r
1 cm
1 cm
!d1 = 1.2 km "
!d2 = 3.1 km "
0.3 km
0.3 km
= 4.0 cm
= 10.3 cm
!
!
Using a ruler and protractor, draw the two vectors, placing the tail of !d2 at the tip of !d1 .
!
!
!
Draw the total displacement vector !dT from the tail of !d1 to the tip of !d2 . Measure the
length of the vector, and measure the angle the displacement vector makes to the horizontal.
The measured length of the total displacement vector is 7.7 cm. Convert the length to kilometres.
!
0.3 km
!dT = 7.7 cm "
= 2.3 km
1 cm
The measured angle ! is 34° .
Statement: The total displacement is 2.3 km [E 34° N].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-1
!
!
2. Given: !d1 = 77.0 m [E]; !d2 = 95.0 m [S]
!
!
Required: !dT ; the angle for !dT , !
!
!
!
Analysis: !dT = !d1 + !d2 . To determine the magnitude of the displacement, use the
Pythagorean theorem. To calculate angle ! , use the tangent ratio.
!
Solution: Solve for !dT .
!
! 2
! 2
!dT = !d1 + !d2
= (77.0 m)2 + (95.0 m)2
!
!dT = 122 m
Solve for angle ! .
r
"d2
tan ! = r
"d1
95.0 m
77.0 m
= 1.2338
! = 51.0°
Statement: The boater’s total displacement is 122 m [E 51.0° S].
!
!
3. Given: !d1 = 65 km [N 32° E]; !d2 = 42 km [E 21° N]
!
!
Required: !dT ; the angle for !dT , !
!
!
!
Analysis: !dT = !d1 + !d2 . To determine the magnitude of the displacement, use the cosine law.
To calculate the angle ! , use the sine law.
Solution: Make a sketch of the addition of displacement vectors.
=
! 2 = 21° + 90° + 32°
! 2 = 143°
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-2
From the cosine law:
a 2 = b2 + c 2 ! 2bccos A
! 2
! 2
! !
! 2
"dT = "d1 + "d2 ! 2 "d1 "d2 cos# 2
= (65 km)2 + (42 km)2 ! 2(65 km)(42 km)(cos 143°)
(three extra digits carried)
! = 101.73 km
2
"dT = 1.0 $ 10 km
From the sine law:
sinC sin A
=
c
a
sin ! 3 sin ! 2
! = !
"d2
"dT
!
"d2 sin ! 2
!
sin ! 2 =
"dT
=
(42 km )(sin 143°)
101.73 km
= 0.24846
! 3 = sin #1 0.24846
! 3 = 14.39° (two extra digits carried)
! = 90° " 32° " 14.39°
! = 44°
Statement: The helicopter travels 1.0 ! 102 km [E 44° N].
Tutorial 2 Practice, page 26
!
1. (a) Given: !d = 25.0 km [E 45.0° N]
Required: Δdx; Δdy
Analysis: Draw the displacement vector, and then use trigonometry to determine the
components. Use east and north as positive.
Solution:
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-3
!
!d x = + !d cos"
!
!d y = + !d sin "
= +(25.0 km)(sin 45.0°)
= +(25.0 km)(cos 45.0°)
!d y = +17.7 km
!d x = +17.7 km
Statement: The components of the displacement are !d x = 17.7 km [E] and !d y = 17.7 km [N] .
r
(b) Given: !d = 355 km [N 42.0° W]
Required: Δdx; Δdy
Analysis: Draw the displacement vector, and then use trigonometry to determine the
components. Use east and north as positive.
Solution:
!
!d x = " !d sin #
= "(355 km)(sin 42.0°)
!d x = "238 km
!
!d y = + !d cos"
= +(355 km)(cos 42.0°)
!d y = +264 km
Statement: The components of the displacement are !d x = 238 km [W] and !d y = 264 km [N] .
!
(c) Given: !d = 32.3 m [E 27.5° S]
Required: Δdx; Δdy
Analysis: Draw the displacement vector, and then use trigonometry to determine the
components. Use east and north as positive.
Solution:
!
!d x = + !d cos"
= +(32.3 m)(cos 27.5°)
!d x = +28.7 m
!
!d y = " !d sin #
= "(32.3 m)(sin 27.5°)
!d y = "14.9 m
Statement: The components of the displacement are !d x = 28.7 m [E] and !d y = 14.9 m [S] .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-4
!
(d) Given: !d = 125 km [S 31.2° W]
Required: Δdx; Δdy
Analysis: Draw the displacement vector, and then use trigonometry to determine the
components. Use east and north as positive.
Solution:
!
!d x = " !d sin #
!
!d y = " !d cos#
= "(125 km)(cos 31.2°)
= "(125 km)(sin 31.2°)
!d x = "64.8 km
!d y = "107 km
Statement: The components of the displacement are !d x = 64.8 km [W] and !d y = 107 km [S] .
Tutorial 3 Practice, page 28
!
!
1. Given: !d1 = 276.9 km [W 76.70° S]; !d2 = 675.1 km [W 11.45° S]
!
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to calculate the x- and y-components of the total
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the total
displacement vector. Use east and north as positive.
Solution: For the first vector,
r
!d1x = " !d1 cos#
= "(276.9 km)(cos 76.70°)
!d1x = "63.701 km (one extra digit carried)
r
!d1y = " !d1 sin #
= "(276.9 km)(sin 76.70°)
!d1y = "269.473 km (two extra digits carried)
For the second vector,
!
!d2x = " !d2 cos#
= "(675.1 km)(cos 11.45°)
!d2x = "661.664 km (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-5
!
!d2y = " !d2 sin #
= "(675.1 km)(sin 11.45°)
!d2y = "134.016 km (two extra digits carried)
Add the horizontal components.
!d x = !d1x + !d2x
= "63.701 km + ("661.664 km)
!d x = "725.365 km
Add the vertical components.
!d y = !d1y + !d2y
= "269.473 km + ("134.016 km)
!d y = "403.489 km
Combine the total displacement components to determine the total displacement.
r
!d = !d x2 + !d y2
r
!d = ("725.365 km)2 + ("403.489 km)2
r
!d = 830.0 km
$ #d y '
! = tan "1 &
)
% #d x (
$ 403.489 km '
= tan "1 &
)
% 725.365 km (
! = 29.09°
Statement: The total displacement of the airplane is 830.0 km [W 29.09° S].
!
!
2. Given: !d1 = 120 km [N 32° W]; !d2 = 150 km [W 24° N]
!
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to calculate the x- and y-components of the total
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the total
displacement vector. Use east and north as positive.
Solution: For the first vector,
!
!d1x = " !d1 sin #
= "(120 km)(sin 32°)
!d1x = "63.59 km (two extra digits carried)
!
!d1y = + !d1 cos"
= +(120 km)(cos 32°)
!d1y = +101.77 km (three extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-6
For the second vector,
!
!d2x = " !d2 cos#
= "(150 km)(cos 24°)
!d2x = "137.03 km (three extra digits carried)
!
!d2y = + !d2 sin "
= +(150 km)(sin 24°)
!d2y = +61.01 km (two extra digits carried)
Add the horizontal components.
!d x = !d1x + !d2x
= "63.59 km + ("137.03 km)
!d x = "200.62 km
Add the vertical components.
!d y = !d1y + !d2y
= +101.77 km + 61.01 km
!d y = +162.78 km
Combine the total displacement components to determine the total displacement.
!
!d = !d x2 + !d y2
= (200.59 km)2 + (162.78 km)2
!
!d = 260 km
$ #d '
y
! = tan &
)
&% #d )(
x
"1
$ 162.78 km '
= tan "1 &
% 200.59 km )(
! = 39°
Statement: The total displacement of the trip is 260 km [W 39° N].
!
!
!
3. Given: !d1 = 12 km [N]; !d2 = 14 km [N 22° E]; !d3 = 11 km [E]
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to calculate the x- and y-components of the total
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the total
displacement vector. Use east and north as positive.
Solution: For the first vector,
!d1x = 0 km
!d1y = +12 km
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-7
For the second vector,
!
!d2x = + !d2 sin "
= +(14 km)(sin 22°)
!d2x = +5.244 km (two extra digits carried)
!
!d2y = + !d2 cos"
= +(14 km)(cos 22°)
!d2y = +12.98 km (two extra digits carried)
For the third vector,
!d3x = 11 km
!d3y = 0 km
Add the horizontal components.
!d x = !d1x + !d2x + !d3x
= 0 km + 5.244 km + 11 km
!d x = +16.244 km
Add the vertical components.
!d y = !d1y + !d2y + !d3y
= +12 km + 12.98 km + 0 km
!d y = +24.98 km
Combine the total displacement components to determine the total displacement.
!
!d = !d x2 + !d y2
= (16.244 km)2 + (24.98 km)2
!
!d = 3.0 " 101 km
!
$ #d '
y
! = tan "1 & ! )
&% #d x )(
$ 24.98 km '
= tan "1 &
% 16.244 km )(
! = 57°
Statement: The total displacement of the helicopter is 3.0 ! 101 km [E 57° N].
Section 1.3 Questions, page 29
!
!
1. (a) Given: !d1 = 7.81 km [E 50° N]; !d2 = 5.10 km [W 11° N]
!
!
Required: !d ; the angle for !d , !
!
!
!
Analysis: !d = !d1 + !d2 Decide on a scale and then draw each vector to scale on a coordinate
axis. Draw the total displacement vector.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-8
Solution: An appropriate scale is 1 cm : 1 km. Calculate the lengths of the arrows for the
displacement:
! 5.10 km
! 7.81 km
= 7.81 cm; !d2 =
= 5.10 cm
!d1 :
1 km /cm
1 km /cm
!
!
Using a ruler and protractor, draw the two vectors placing the tail of !d2 at the tip of !d1 . Draw
!
!
!
the total displacement vector !d from the tail of !d1 to the tip of !d2 . Measure the length of the
vector, and measure the angle the displacement vector makes to the horizontal.
The measured length of the total displacement vector is 7.0 cm, measured to the nearest
millimetre. Convert to kilometres.
r
1 km
!d = 7.0 cm "
1 cm
= 7.0 km
The measured angle ! is 0° , measured to the nearest degree.
Statement: The total displacement is 7.0 km [N].
(b) Both an algebraic solution and a component solution are given.
(i) Algebraic!Method:
!
Required: !d ; the angle for !d , !
!
!
!
Analysis: !d = !d1 + !d2 . To determine the magnitude of the displacement, use the cosine law.
To calculate the angle ! , use the sine law.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-9
Solution: Make a sketch of the addition of displacement vectors.
The angle ! 2 is
! 2 = 50° + 11°
! 2 = 61°
From the cosine law,
a 2 = b2 + c 2 ! 2bccos A
! 2
! 2
! !
!2
"d = "d1 + "d2 ! 2 "d1 "d2 cos# 2
= (7.81 km)2 + (5.10 km)2 ! 2(7.81 km)(5.10 km)(cos 61°)
!2
"d = 48.385 m 2
!
"d = 6.956 m (two extra digits carried)
!
"d = 7.0 m
From the sine law,
sinC sin A
=
c
a
sin ! 3 sin ! 2
! =
!
"d2
"d
!
"d2 sin ! 2
!
sin ! 3 =
"d
=
(5.10 m )(sin 61°)
(6.956 m )
= 0.6412
! 3 = sin #1 0.6412
! 3 = 39.88°
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-10
! = 50° + 39.88°
! = 90°
Statement: The total displacement is 7.0 km [N].
(ii) Component
! Method:
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to calculate the x- and y-components of the total
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the total
displacement vector. Use east and north as positive.
Solution: For the first vector,
!
!d1x = + !d1 cos"
= +(7.81 km)(cos 50°)
!d1x = #5.020 km (two extra digits carried)
!
!d1y = + !d1 sin "
= +(7.81 km)(sin 50°)
!d1y = +5.983 km (two extra digits carried)
For the second vector,
!
!d2x = " !d2 cos#
= "(5.10 km)(cos 11°)
!d2x = "5.006 km (two extra digits carried)
!
!d2y = + !d2 sin "
= +(5.10 km)(sin 11°)
!d2y = +0.9732 km (two extra digits carried)
Add the horizontal components.
!d x = !d1x + !d2x
= +5.020 km + ("5.006 km)
!d x = +0.014 km
Add the vertical components.
!d y = !d1y + !d2y
= +5.983 km + 0.9732 km
!d y = +6.9562 km
Combine the total displacement components to determine the total displacement.
!
!d = !d x2 + !d y2
=
(0.014 km ) + (6.9562 km )
2
2
!
!d = 7.0 km
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-11
$ #d '
y
! = tan "1 &
)
&% #d x )(
$ 6.9562 km '
= tan "1 &
)
% 0.014 km (
! = 90°
Statement: The total displacement of the trip is 7.0 km [N].
(c) Answers may vary. Sample answer: The answers obtained by the three methods were the
same (no percent difference). It is reasonable that the methods in part (b) gave the same results
because care was taken to carry extra digits while calculating. In both cases, the final answer was
given to two significant digits, consistent with the given information. It is somewhat surprising
that the result of part (a) agreed with the mathematical methods because it is difficult to draw or
measure an angle to 1° precision. It is also difficult to draw arrows tip to tail to 1 mm precision.
!
!
!
2. Given: !d1 = 5.0 cm [E 30.0° N]; !d2 = 7.5 cm [E]; !d3 = 15.0 cm [E 10.0° S]
!
!
Required: !d ; the angle for !d , !
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Decide on a scale and then draw each vector to scale on a
coordinate axis. Draw the total displacement vector.
Solution: An appropriate scale is 1 cm : 2 cm. The lengths of the arrows for the displacement
vectors are
!
1 cm
!d1 = 5.0 cm "
= 2.5 cm
2 cm
!
1 cm
!d2 = 7.5 cm
= 3.75 cm
2 cm
!
1 cm
!d3 = 15.0 cm
= 7.5 cm
2 cm
!
!
Using a ruler and protractor, draw the three vectors placing the tail of !d2 at the tip of !d1 , and
!
!
!
!
the tail of !d3 at the tip of !d2 . Draw the total displacement vector !d from the tail of !d1 to
!
the tip of !d3 . Measure the length of the displacement vector, and measure the angle the
displacement vector makes to the horizontal.
(
(
)
)
The measured length of the total displacement vector is 13.5 cm, to the nearest millimetre.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-12
Convert back to the original scale.
!
2 cm
= 27 cm
!d = 13.5 cm "
1 cm
The measured angle ! is 0° , to the nearest degree.
Statement: The total displacement is 27 cm [E].
!
3. Given: !d = 2.50 m [N 38.0° W]
Required: !d x ; !d y
Analysis:
Draw the displacement vector, and then use trigonometry to determine the components. Use east
and north as positive.
!
!
Solution: !d x = " !d sin #
!d y = + !d cos"
= "(2.50 m)(sin 38.0°)
!d x = "1.54 m
= +(2.50 km)(cos 38.0°)
!d y = +1.97 m
Statement: The components of the displacement are !d x = 1.54 m [W] and !d y = 1.97 m [N] .
!
4. Given: !d = 25.0 m [E 30.0° N]
Required: !d x ; !d y
Analysis: Draw the displacement vector, and then use trigonometry to determine the
components. Use east and north as positive.
Solution:
!
!d x = + !d cos"
= +(25.0 m)(cos 30.0°)
!d x = +21.6 m
!
!d y = + !d sin "
= +(25.0 m)(sin 30.0°)
!d y = +12.5 m
Statement: The components of the displacement are !d x = 21.6 m [E] and !d y = 12.5 m [N] .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-13
5. Given: !d x = 54 m [E]; !d y = 24 m [N]
!
Required: !d
Analysis: Use the Pythagorean theorem to determine the length of the displacement vector. Use
east and north as positive.
!
Solution: !d = !d x2 + !d y2
= (54 m ) + ( 24 m )
2
2
!
!d = 59 m
Statement: The length of the displacement vector is 59 m.
(b) Required: !
Analysis: Use the tangent ratio to calculate ! .
!
$ #d '
y
Solution: ! = tan "1 & ! )
&% #d x )(
$ 24 m '
= tan "1 &
% 54 m )(
! = 24°
Statement: The vector points [E 24° N].
!
!
!
6. Given: !d1 = 15.0 km [W]; !d2 = 45.0 km [S]; !d3 = 32 km [N 25° W]
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to calculate the x- and y-components of the total
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the total
displacement vector. Use east and north as positive.
Solution: For the first vector,
!d1x = "15.0 km
!d1y = 0 km
For the second vector,
!d2 x = 0 km
!d2y = "45.0 km
For the third vector,
!
!d3x = " !d3 sin #
= "(32 km)(sin 25°)
!d3x = "13.52 km (two extra digits carried)
!
!d3y = + !d3 cos"
= +(32 km)(cos 25°)
!d3y = +29.00 km (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-14
Add the horizontal components.
!d x = !d1x + !d2x + !d3x
= ("15.0 km) + 0 km + ("13.52 km)
!d x = "28.52 km
Add the vertical components.
!d y = !d1y + !d2y + !d3y
= 0 km + ("45.0 km) + 29.00 km
!d y = "16.00 km
Combine the total displacement components to determine the total displacement.
!
!d = !d x2 + !d y2
=
( 28.52 km )2 + (16.00 km )2
!
!d = 33 km
!
$ #d
y
! = tan "1 & !
&% #d x
'
)
)(
$ 16.00 km '
= tan "1 &
)
% 28.52 km (
! = 29°
Statement: The total displacement of the driver is 33 km [W 29° S].
!
!
!
7. Given: !d1 = 2.5 m [W 30.0° S]; !d2 = 3.6 m [S]; !d3 = 4.9 m [E 38.0° S]
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to calculate the x- and y-components of the total
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the total
displacement vector. Use east and north as positive.
Solution: For the first vector,
!
!d1x = " !d1 cos#
= "(2.5 m)(cos 30.0°)
!d1x = "2.165 m (two extra digits carried)
!
!d1y = " !d1 sin #
= "(2.5 m)(sin 30.0°)
!d1y = "1.250 m (two extra digits carried)
For the second vector,
!d2 x = 0 m
!d2y = "3.6 m
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-15
For the third vector,
!
!d3x = + !d3 cos"
= +(4.9 m)(cos 38.0°)
!d3x = +3.861 m (two extra digits carried)
!
!d3y = " !d3 sin #
= "(4.9 m)(sin 38.0°)
!d3y = "3.017 m (two extra digits carried)
Add the horizontal components.
!d x = !d1x + !d2x + !d3x
= ("2.165 m) + 0 m + 3.861 m
!d x = 1.696 m
Add the vertical components.
!d y = !d1y + !d2y + !d3y
= "1.250 m + ("3.6 m) + ("3.017 m)
!d y = "7.867 m
Combine the components to determine the total displacement vector.
!
!d = !d x2 + !d y2
=
(1.696 m )2 + ( 7.867 m )2
!
!d = 8.0 m
!
$ #d
y
! = tan "1 & !
&% #d x
'
)
)(
$ 7.867 m '
= tan "1 &
% 1.696 m )(
! = 78°
Statement: The total displacement is 8.0 m [E 78° S].
!
!
8. (a) Given: !d1 = 2.70 km [E 25.0° N]; !d2 = 4.80 km [E 45.0° S]
Required: !d x ; !d y
!
!
!
Analysis: !d = !d1 + !d2 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to calculate the x- and y-components of the total
displacement. Use east and north as positive.
Solution: For the first vector,
!
!d1x = + !d1 cos"
= +(2.70 km)(cos 25.0°)
!d1x = +2.4470 km (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-16
!
!d1y = + !d1 sin "
= +(2.70 km)(sin 25.0°)
!d1y = +1.1411 km (two extra digits carried)
For the second vector,
!
!d2x = + !d2 cos"
= +(4.80 km)(cos 45.0°)
!d2x = +3.3941 km (two extra digits carried)
!
!d2y = " !d2 sin #
= "(4.80 km)(sin 45.0°)
!d2y = "3.3941 km (two extra digits carried)
Add the horizontal components.
!d x = !d1x + !d2x
= +2.4470 km + 3.3941 km
= +5.841 km
!d x = +5.84 km
Add the vertical components.
!d y = !d1y + !d2y
= +1.1411 km + ("3.3941 km)
= "2.2530 km
!d y = "2.25 km
Statement: The components of the boat’s displacement are !d x = 5.84 km [E]
and !d y = 2.25 km [S] .
!
(b) Required: !d
Analysis: Use the Pythagorean theorem and tangent ratio to determine the total displacement
vector.
!
!d = !d x2 + !d y2
=
(5.8411 km )2 + ( 2.2530 km )2
!
!d = 6.26 km
$ #d '
y
! = tan "1 &
)
&% #d x )(
$ 2.2530 km '
= tan "1 &
)
% 5.8411 km (
! = 21.1°
Statement: The total displacement of the boat is 6.26 km [E 21.1° S].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-17
!
!
9. Given: !d1 = 1512.0 km [W 19.30° N]; !d2 = 571.0 km [W 4.35° N];
!
!d3 = 253.1 km [W 39.39° N]
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to calculate the x- and y-components of the total
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the total
displacement vector. Notice that all of the displacements point somewhat west and north. Use
west and north as positive.
Solution: For the first vector,
!
!d1x = !d1 cos"
= (1512.0 km)(cos 19.30°)
!d1x = 1427.027 km (two extra digits carried)
!
!d1y = !d1 sin "
= (1512.0 km)(sin 19.30°)
!d1y = 499.738 km (two extra digits carried)
For the second vector,
!
!d2x = !d2 cos"
= (571.0 km)(cos 4.35°)
!d2x = 569.355 km (two extra digits carried)
!
!d2y = !d2 sin "
= (571.0 km)(sin 4.35°)
!d2y = 43.310 km (two extra digits carried)
For the third vector,
!
!d3x = !d3 cos"
= (253.1 km)(cos 39.39°)
!d3x = 195.607 km (two extra digits carried)
!
!d3y = !d3 sin "
= (253.1 km)(sin 39.39°)
!d3y = 160.616 km (two extra digits carried)
Add the horizontal components.
!d x = !d1x + !d2x + !d3x
= 1427.027 km + 569.355 km + 195.607 km
!d x = 2191.989 km
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-18
Add the vertical components.
!d y = !d1y + !d2y + !d3y
= 499.738 km + 43.310 km + 160.616 km
!d y = 703.664 km
Combine the components to determine the total displacement vector.
!
!d = !d x2 + !d y2
=
( 2191.989 km )2 + ( 703.664 m )2
= 2302.164 km
!
!d = 2.30 " 103 km
$ #d '
y
! = tan "1 &
)
&% #d x )(
$ 703.664 km '
= tan "1 &
)
% 2191.989 km (
! = 17.8°
Statement: The total displacement is 2.30 ! 103 km [W 17.8° N].
r
r
10. Given: !d1 = 25 km [N]; !dT = 62 km [N 38° W]
!
Required: !d2
r
r
r
Analysis: !dT = !d1 + !d2 . Determine the x- and y-components of the given displacement
vectors. Subtract these x- and y-components to calculate the x- and y-components of the second
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the second
displacement vector. Use east and north as positive.
Solution: For the first vector:
!d1x = 0 km
!d1y = 25 km
For the total vector,
!
!d x = " !d sin #
= "(62 km)(sin 38°)
!d x = "38.171 km (two extra digits carried)
!
!d y = !d cos"
= (62 km)(cos 38°)
!d y = 48.857 km (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-19
Subtract the horizontal components.
!d2x = !d x " !d1x
= ("38.171 km) + (0 km)
!d2x = "38.171 km
Subtract the vertical components.
!d2y = !d y " !d1y
= (48.857 km) + (25 km)
!d2y = 23.857 km
Combine the displacement components of the second vector to determine the second
displacement.
!
!d2 = !d2x2 + !d2y2
=
( "38.171 km )2 + ( 23.857 km )2
!
!d2 = 45 km
!
$ #d
2y
! = tan "1 & !
&% #d2x
'
)
)(
$ 23.857 km '
= tan "1 &
)
% 38.171 km (
! = 32°
Statement: The second displacement is 45 km [W 32° N].
11. Answers may vary. Sample answer: The length of the total displacement vector is determined
by how the two displacement vectors line up.
If the second vector points west, parallel to the first vector, then the total displacement is as large
as possible: 450 km [W] + 220 km [W] = 670 km [W]
If the second vector points east, opposite to the first vector, then the total displacement is as
small as possible: 450 km [W] + 220 km [E] = 230 km [W]
If the second vector points in any other direction, the magnitude of the total displacement is
between 230 km and 670 km.
12. Answers may vary. Sample answer: Vector addition is commutative. This means that it does
not matter what order you use when you add two vectors. You can put the tail of the second to
the tip of the first, or you can put the tail of the first to the tip of the second. You get the same
total vector either way.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.3-20
Section 1.4: Velocity and Acceleration in Two Dimensions
Tutorial 1 Practice, page 32
!
!
!
1. (a) Given: !d1 = 72.0 km [W 30.0° S]; !d2 = 48.0 km [S]; !d3 = 150.0 km [W]; !t = 2.5 h
!
Required: !d
!
!
!
!
Analysis: !d = !d1 + !d2 + !d3 . Determine the x- and y-components of the displacement
vectors. Then determine the x- and y-components of the total displacement using
!d x = !d1x + !d2x + !d3x and !d y = !d1y + !d2y + !d3y . Finally, combine these components to
determine the total displacement vector using the Pythagorean theorem and the tangent ratio.
Use east and north as positive.
!
Solution: x-component of !d :
!d x = !d1x + !d2x + !d3x
= "(72.0 km)(cos 30.0°) + 0 km + ("150.0 km)
= "62.35 km + 0 km + ("150.0 km)
!d x = "212.4 km (two extra digits carried)
!
y-component of !d :
!d y = !d1y + !d2y + !d3y
= "(72.0 km)(sin 30.0°) + ( "48.0 km ) + 0 km
= "36.0 km + ("48.0 km) + 0 km
!d y = "84.0 km (one extra digit carried)
Combine the total displacement components to determine the total displacement.
!
!d = !d x2 + !d y2
= ("212.4 km)2 + ("84.0 km)2
= 228.4 km (two extra digits carried)
!
!d = 230 km
$ #d '
y
! = tan &
)
&% #d x )(
"1
$ 84.0 km '
= tan "1 &
)
% 212.4 km (
! = 22°
Statement: The total displacement is 230 km [W 22° S].
!
(b) Given: !d = 230 km [W 22º S]; !t = 2.5 h
!
Required: vav
!
!d
!
Analysis: Use vav =
to determine the average velocity.
!t
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-1
!
!d
!
Solution: vav =
!t
228.4 km [W 22° S]
=
2.5 h
!
vav = 91 km/h [W 22° S]
Statement: The average velocity of the plane is 91 km/h [W 22° S].
!
!
!
(c) Given: !d1 = 72.0 km [W 30.0° S]; !d2 = 48.0 km [S]; !d3 = 150.0 km [W]; !t = 2.5 h
Required: vav
Analysis: Calculate the total distance travelled and then use vav =
speed.
Solution: !d = !d1 + !d2 + !d3
= 72.0 km + 48.0 km + 150.0 km
!d = 270.0 km (two extra digits carried)
!d
to determine the average
!t
!d
!t
270.0 km
=
2.5 h
= 108 km/h
vav =
vav = 110 km/h
Statement: The average speed of the plane is 110 km/h.
!
!
!
2. (a) Given: !d1 = 25.0 km [E 53.13° N]; !d2 = 20.0 km [S]; !d3 = 15.0 km [W]; !t = 12 h
!
Required: vav
!
!
!
!
Analysis: Calculate the total displacement using components and !d = !d1 + !d2 + !d3 . Then,
!
!d
!
calculate the average velocity using vav =
. Use east and north as positive.
!t
!
Solution: x-component of !d :
!d x = !d1x + !d2x + !d3x
= (25.0 km)(cos 53.13°) + 0 km + ("15.0 km)
= 15.0 km + 0 km + ("15.0 km)
!d x = 0 km
!
y-component of !d :
!d y = !d1y + !d2y + !d3y
= (25.0 km)(sin 53.13°) + ("20.0 km) + 0 km
= 20.0 km + ("20.0 km) + 0 km
!d y = 0 km
!
Both components of the total displacement are 0 km. Hence the total displacement is !d = 0 km .
As a result, the average velocity is 0 km/h.
Statement: The average velocity of the elk is 0 km/h.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-2
(b) Given: !d1 = 25.0 km; !d2 = 20.0 km; !d3 = 15.0 km; !t = 12 h
Required: vav
Analysis: Calculate the total distance travelled and then use vav =
speed.
Solution: !d = !d1 + !d2 + !d3
= 25.0 km + 20.0 km + 15.0 km
!d = 60.0 km (one extra digit carried)
!d
to determine the average
!t
!d
!t
60.0 km
=
18 h
vav = 5.0 km/h
vav =
Statement: The average speed of the elk is 5.0 km/h.
(c) Answers may vary. Sample answer: While the elk travelled 60.0 km, it returned to its starting
place. Average velocity depends on displacement, a vector. From a vector point of view, the elk
did not move, resulting in an average velocity of 0 km/h.
Tutorial 2 Practice, page 34
!
!
1. Given: vi = 20.0 m/s [E]; vf = 20.0 m/s [S]; !t = 12 s
!
Required: aav
Analysis: Draw a vector diagram of the situation. Calculate the change in velocity using
!
!v
!
! ! !
. Use east
components and !v = vf " vi . Then, determine the average acceleration using aav =
!t
and north as positive.
Solution:
!
x-component of !v :
!vx = vf x " vi x
= 0 m/s " 20.0 m/s
!vx = "20.0 m/s
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-3
!
y-component of !v :
!v y = vf y " vi y
= ("20 m/s) " 0 m/s
!v y = "20.0 m/s
Determine the change in velocity from its components.
!
!v = !vx2 + !v 2y
=
( "20.0 m/s )2 + ( "20.0 m/s )2
!
!v = 28.28 m/s (two extra digits carried)
$ #v '
y
! = tan "1 &
)
&% #vx )(
$ 20.0 m/s '
= tan "1 &
)
% 20.0 m/s (
! = 45°
Calculate the average acceleration.
!
!v
!
aav =
!t
28.28 m/s [W 45° S]
=
12 s
!
2
aav = 2.4 m/s [W 45° S]
Statement: The average acceleration of the car is 2.4 m/s 2 [W 45° S] .
!
!
2. Given: vi = 50.0 km/h [W 60.0° N]; vf = 80.0 km/h [E 60.0° N]; !t = 15.0 min = 0.250 h
!
Required: aav
Analysis: Draw a vector diagram of the situation. Calculate the change in velocity using
!
!v
!
! ! !
components and !v = vf " vi . Then, determine the average acceleration using aav =
. Use east
!t
and north as positive.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-4
Solution:
!
The x-component of !v is
!vx = vf x " vi x
= (80.0 km/h)(cos 60.0°) " ("(50.0 km/h)cos 60.0°)
= 40.0 km/h " ("25.0 km/h)
!vx = 65.0 km/h
!
The y-component of !v is
!v y = vf y " vi y
= (80.0 km/h)(sin 60.0°) " (50.0 km/h)(sin 60.0°)
= 69.28 km/h " 43.30 km/h
!v y = 25.98 km/h (one extra digit carried)
Determine the change in velocity from its components.
!
!v = !vx2 + !v 2y
=
(65.0 km/h )2 + ( 25.98 km/h )2
!
!v = 70.00 km/h (one extra digit carried)
$ #v '
y
! = tan "1 &
)
&% #vx )(
$ 25.98 km/h '
= tan "1 &
)
% 65.0 km/h (
! = 21.8°
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-5
Calculate the average acceleration.
!
!v
!
aav =
!t
70.00 km/h [E 21.8° N]
=
0.250 h
!
aav = 2.80 " 102 km/h 2 [E 21.8° N]
Statement: The average acceleration of the truck is 2.80 ! 102 km/h 2 [E 21.8° N] .
!
!
3. (a) Given: !d1 = 800.0 km [E 7.5° S]; !d2 = 400.0 km [E 51° S]; !t = 18.0 h
Required: !d
Analysis: Calculate the total distance travelled by adding the individual distances
travelled, !d = !d1 + !d2 .
Solution: !d = !d1 + !d2
= 800.0 km + 400.0 km
!d = 1.2 " 103 km
Statement: The total distance travelled by the bird is 1.2 × 103 km.
!
!
(b) Given: !d1 = 800.0 km [E 7.5° S]; !d2 = 400.0 km [E 51° S]
!
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 . Determine the x- and y-components of the displacement vectors.
Then determine the x- and y-components of the total displacement using !d x = !d1x + !d2x
and !d y = !d1y + !d2y . Finally, combine these components to determine the total displacement
vector using the Pythagorean theorem and the tangent ratio. Use east and north as positive.
!
Solution: The x-component of !d is
!d x = !d1x + !d2x
= (800.0 km)(cos 7.5°) + (400.0 km)(cos 51°)
= 793.1 km + 251.7 km
!d x = 1045 km (two extra digits carried)
!
The y-component of !d is:
!d y = !d1y + !d2y
= "(400.0 km)(sin 7.5°) + ("(800.0 km)sin 51°)
= "104.4 km + ("310.9 km)
!d y = "415.3 km (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-6
Combine the total displacement components to determine the total displacement.
!
!d = !d x2 + !d y2
= ("1045 km)2 + ("415.3 km)2
= 1124 km
!
!d = 1.1# 103 km
$ #d '
y
! = tan "1 &
)
&% #d x )(
$ 415.3 km '
= tan "1 &
)
% 1045 km (
! = 22°
Statement: The total displacement is 1.1! 103 km [E 22° S].
(c) Given: !t = 18.0 h; !d = 1200 km
Required: vav
Analysis: Use vav =
!d
to determine the average speed.
!t
!d
!t
1200 km
=
18.0 h
vav = 67 km/h
Statement: The average speed of the bird is 67 km/h.
r
(d) Given: !t = 18.0 h; !d = 1124 km [E 22o S]
!
Required: vav
!
!d
!
Analysis: Use vav =
to determine the average velocity.
!t
r
r
!d
Solution: vav =
!t
1124 km [E 22° S]
=
18.0 h
r
vav = 62 km/h [E 22° S]
Statement: The average velocity of the bird is 62 km/h [E 22° S] .
Solution: vav =
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-7
Section 1.4 Questions, page 35
1. Answers may vary. Sample answer: The average speed is based on the distance travelled.
Changes in direction do not affect the average speed. The average velocity is based on the
displacement. The displacement is affected by changes in direction: it acts like the short-cut
distance from start to finish. For example, a dog walks 40 m [E] and then 20 m [W] in 60 s. It
travels a total of 60 m, giving an average speed of 1 m/s. The dog’s displacement is 20 m [E]
because it backtracked. You do not see the individual displacements of its walk. The dog’s
average velocity is 0.33 m/s [E]. Based just on the start and finish, the dog appears to move quite
slowly.
!
!
2. (a) Given: !d1 = 25.0 m [E 30.0° N]; !d2 = 75.0 km [E 45.0° S]; !t = 4.0 min = 240 s
Required: !d
Analysis: Calculate the total distance travelled by adding the individual distances
travelled, !d = !d1 + !d2 .
Solution: !d = !d1 + !d2
= 25.0 m + 75.0 m
!d = 1.0 " 102 m
Statement: The total distance travelled by the loon is 1.0 × 102 m.
!
!
(b) Given: !d1 = 25.0 m [E 30.0° N]; !d2 = 75.0 km [E 45.0° S]
!
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 ; Determine the x- and y-components of the displacement vectors.
Then determine the x- and y-components of the total displacement using !d x = !d1x + !d2x
and !d y = !d1y + !d2y . Finally, combine these components to determine the total displacement
vector using the Pythagorean theorem
and the tangent ratio. Use east and north as positive.
!
Solution: The x-component of !d is
!d x = !d1x + !d2x
= (25.0 m)(cos 30.0°) + (75.0 m)(cos 45.0°)
= 21.6506 m + 53.0330 m
!d x = 74.684 m (two extra digits carried)
!
The y-component of !d is
!d y = !d1y + !d2y
= (25.0 m)(sin 30.0°) + ("(75.0 m)sin 45.0°)
= 12.5 m + ( "53.0330 m )
!d y = "40.533 m (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-8
Combine the total displacement components to determine the total displacement.
r
!d = !d x2 + !d y2
= (74.684 m)2 + ("40.533 m)2
= 84.974 m (two extra digits carried)
r
!d = 85.0 m
$ #d '
y
! = tan &
)
&% #d x )(
"1
$ 40.533 km '
= tan "1 &
)
% 74.684 km (
! = 28.5°
Statement: The loon’s total displacement is 85.0 m [E 28.5° S].
(c) Given: !d = 1.0 " 102 m; !t = 4.0 min = 240 s
Required: vav
Analysis: Use vav =
!d
to determine the average speed.
!t
!d
!t
1.0 " 102 m
=
240 s
vav = 0.42 m/s
Solution: vav =
Statement: The average speed of the loon is 0.42 m/s.
!
(d) Required: vav
!
!d
!
Analysis: Use vav =
to determine the average velocity.
!t
!
!d
!
Solution: vav =
!t
84.9739 m [E 28.5° S]
=
240 s
!
vav = 0.35 m/s [E 28.5° S]
Statement: The average velocity of the loon is 0.35 m/s [E 28.5° S] .
!
!
3. (a) Given: !d1 = 15.0 km [W 30.0° N]; !d2 = 10.0 km [W 75.0° N];
!
!d3 = 10.0 km [E 70.0° N]; !t = 0.50 h
!
(b) Required: vav
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-9
!
!
!
!
Analysis: Calculate the total displacement using components and !d = !d1 + !d2 + !d3 . Then,
!
!d
!
calculate the average velocity using vav =
. Use east and north as positive.
!t
!
Solution: The x-component of !d is
!d x = !d1x + !d2x + !d3x
= "(15.0 km)(cos 30.0°) + ("(10.0 km)cos 75.0°) + (10.0 km)(cos 70.0°)
= "12.990 km + ( "2.588 km ) + 3.420 km
!d x = "12.158 km (two extra digits carried)
!
The y-component of !d is
!d y = !d1y + !d2y + !d3y
= +(15.0 km)(sin 30.0°) + (10.0 km)(sin 75.0°) + (10.0 km)(sin 70.0°)
= +7.5 km + 9.659 km + 9.397 km
!d y = +26.556 km (two extra digits carried)
Combine the total displacement components to determine the total displacement.
!
!d = !d x2 + !d y2
= ("12.158 km)2 + (26.556 km)2
!
!d = 29.207 km
$ #d '
y
! = tan "1 &
)
&% #d x )(
$ 26.556 km '
= tan "1 &
)
% 12.158 km (
! = 65.4°
The average
velocity is
!
!d
!
vav =
!t
29.207 km [W 65.4° N]
=
0.50 h
!
vav = 58 km/h [W 65.4° N]
Statement: The driver’s average velocity is 58 km/h [W 65.4° N] .
4. Answers may vary. Sample answer: Acceleration results from a change in velocity. The
change may involve a change in speed or a change in direction. Going around a curve at constant
speed is a situation where there is average acceleration and no change in speed.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-10
5. Answers may vary. Sample answers: One way to subtract vectors is to use components.
Determine the components of each vector. Subtract the x-components and then the
y-components. Build the resulting vector out of its components. A second way is similar to
adding vectors using the cosine and sine laws. Draw the first vector. Add to it the negative of the
second vector. Draw the resulting vector to complete the triangle. Then solve the triangle.
!
!
6. Given: !d1 = 150 km [E 12° N]; !vav = 130 km/h [N 32° E]; !t1 = 1.0 h; !t = 3.0 h
!
Required: !d2
!
!d
!
Analysis: Use vav =
to calculate the total displacement. Then determine the x- and
!t
y-components of the first and total displacement vectors. Subtract these x- and y-components to
!
!
!
determine the x- and y-components of the second displacement, !d = !d1 + !d2 . Finally, use the
Pythagorean theorem and tangent ratio to determine the second displacement vector. Use east
and north as positive.
!
!
d
Solution:
Determine
the
total
displacement
.
!
!d
!
vav =
! !!t
!d = vav !t
= (130 km/h [N 32° E])(3.0 h)
!
!d = 390 km [N 32° E]
!
The x-component of !d2 is
!d2x = !d x " !d1x
= (390 km)(sin 32°) " (150 km)(cos12°)
= 206.67 km " 146.72 km
!d2x = 59.95 km (two extra digits carried)
!
The y-component of !d2 is:
!d2y = !d y " !d1y
= (390 km)(cos 32°) " (150 km)(sin 12°)
= 330.74 km " 31.19 km
!d2y = 299.55 km (three extra digits carried)
Combine the displacement components of the second vector to determine the second
displacement.
!
!d2 = !d2x2 + !d2y2
=
(59.95 km )2 + ( 299.55 km )2
= 305.49 km
!
!d2 = 3.1" 102 km
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-11
$ #d '
2y
! = tan "1 &
)
&% #d2x )(
$ 299.55 km '
= tan "1 &
)
% 59.95 km (
! = 79°
Statement: The second displacement is 3.1! 102 km [N 11° E].
!
!
7. Given: vav = 3.5 m/s; !d1 = 1.8 km [E]; !d2 = 2.6 km [N 35° E]
Required: !t
Analysis: The jogger returns to his starting place, so the total displacement is 0 m,
!
!
!
!d1 + !d2 + !d3 = 0 m . Use vector subtraction by components to determine the magnitude of the
third displacement vector. Use !d = !d1 + !d2 + !d3 to calculate the total distance the jogger
!d
. Use east and north as positive.
ran. Finally determine the time taken using vav =
!t
!
Solution: The x-component of !d3 is
!d1x + !d2x + !d3x = 0 m
!d3x = "!d1x " !d2x
= "(1.8 km) " (2.6 km)(sin35°)
= "(1.8 km) " 1.491 km
!d3x = "3.291 km (two extra digits carried)
!
The y-component of !d3 is
!d1y + !d2y + !d3y = 0 m
!d3y = "!d1y " !d2y
= "(0 km) " (2.6 km)(cos 35°)
= "(0 km) " 2.130 km
!d3y = "2.130 km (two extra digits carried)
Combine these displacement components to determine the magnitude of the third displacement.
!
!d = !d x2 + !d y2
=
( "3.291 km )2 + ( 2.130 km )2
!
!d = 3.920 km
The total distance run is
!d = !d1 + !d2 + !d3
= (1.8 km) + (2.6 km) + (3.920 km)
!d = 8.320 km
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-12
The time taken is
!d
!t =
vav
8.320 km 1000 m
"
3.5 m/s
1 km
= 2377 s
=
!t = 2.4 " 103 s
Statement: The jogger’s run takes 2.4 ! 103 s .
r
r
8. Given: v1 = 6.4 m/s [E 30.0° S]; v2 = 8.5 m/s [E 30.0° N]; !t = 3.8 s
!
Required: aav
! ! !
Analysis: Calculate the change in velocity using components and !v = vf " vi . Then determine
!
!v
!
the average acceleration from aav =
. Use east and north as positive.
!t
!
Solution: The x-component of !v is
!vx = vf x " vi x
= (8.5 m/s)(cos 30.0°) " (6.4 m/s)(cos 30.0°)
= 7.361 m/s " 5.543 m/s
!vx = 1.818 m/s (two extra digits carried)
!
The y-component of !v is
!v y = vf y " vi y
= (8.5 m/s)(sin 30.0°) " ("(6.4 m/s)(sin 30.0°))
= 4.25 m/s " ("3.2 m/s)
!v y = 7.45 m/s (one extra digit carried)
Determine the change in velocity from its components.
!
!v = !vx2 + !v 2y
=
(1.818 m/s )2 + ( 7.45 m/s )2
!
!v = 7.669 m/s
$ #v '
y
! = tan "1 &
)
&% #vx )(
$ 7.45 m/s '
= tan "1 &
)
% 1.818 m/s (
! = 76°
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-13
Calculate the average acceleration.
!
!v
!
aav =
!t
7.669 m/s [E 76° N]
=
3.8 s
!
2
aav = 2.0 m/s [E 76° N]
Statement: The average acceleration of the bird is 2.0 m/s 2 [E 76° N] .
!
!
9. Given: vi = 50.0 m/s [W]; vf = 35.0 m/s [S]; !t = 45.0 s
!
Required: aav
! ! !
Analysis: Draw a vector diagram showing the change in velocity using !v = vf " vi . Solve the
!
!v
!
triangle using trigonometry. Then determine the average acceleration from aav =
. Use east
!t
and north as positive.
Solution:
!
Use the Pythagorean theorem to determine !v .
r
!v =
r2
r
!vi + !vf
2
= (50.0 m/s)2 + (35.0 m/s)2
= 61.033 m/s (two extra digits carried)
!
$ #vf '
"1
! = tan & ! )
% #vi (
$ 35.0 m/s '
= tan "1 &
)
% 50.0 m/s (
! = 35.0°
Calculate the average acceleration.
r
r
!v
aav =
!t
61.033 m/s [E 35.0° S]
=
45.0 s
r
aav = 1.36 m/s 2 [E 35.0° S]
Statement: The average acceleration of the helicopter is 1.36 m/s 2 [E 35.0° S] .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-14
r
r
10. Given: vi = 8.2 m/s [E 25° S]; vf = 8.2 m/s [E 25° N]; !t = 3.2 ms = 3.2 " 10#3 s
!
Required: aav
Analysis: Draw a vector diagram of the situation. Calculate the change in velocity using
!
!v
!
! ! !
components and !v = vf " vi . Then, determine the average acceleration from aav =
. Use east
!t
and north as positive.
Solution: Components for the initial velocity vector:
Components for the final velocity vector:
!
The x-component of !v is
!vx = vf x " vi x
= (8.2 m/s)(cos 25°) " (8.2 m/s)(cos 25°)
= 7.432 m/s " 7.432 m/s
!vx = 0 m/s
!
The y-component of !v is
!v y = vf y " vi y
= (8.2 m/s)(sin 25°) " ("(8.2 m/s)(sin 25°))
= 3.465 m/s " ("3.465 m/s)
!v y = 6.930 m/s (two extra digits carried)
Determine the change in velocity from its components.
r
!v = !vx2 + !v 2y
=
(0 m/s )2 + ( 6.930 m/s )2
r
!v = 6.930 m/s (two extra digits carried)
!
!
Since !vx = 0 m/s , !v points north.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-15
Calculate the average acceleration.
!
!v
!
aav =
!t
6.930 m/s [N]
=
3.2 " 10#3 s
!
aav = 2.2 " 103 m/s 2 [N]
Statement: The average acceleration of the pool ball is 2.2 ! 103 m/s 2 [N] .
!
!
11. Given: vi = 6.4 m/s [W 35° N]; aav = 2.2 m/s 2 [S]; !t = 4.0 s
!
Required: vf
Analysis: Calculate the change in velocity from the acceleration and time interval:
!
!v
!
aav =
!t
! !
!v = aav !t
! ! !
!
Using components and !v = vf " vi , determine vf . Use east and north as positive.
Solution: The change in velocity is
! !
!v = aav !t
= ( 2.2 m/s 2 [S]) (4.0 s)
!
!v = 8.8 m/s [S]
!
The x-component of vf is
vf x = vi x + !vx
= "(6.4 m/s)(cos 35°) + 0 m/s
= "5.246 m/s + 0 m/s
vf x = "5.246 m/s
!
The y-component of vf is
vf y = viy + !v y
= (6.4 m/s)(sin 35°) + ( "8.8 m/s )
= 3.671 m/s + ("8.8 m/s)
vf y = "5.129 m/s
Determine the final velocity from its components.
2
2
r
vf = vfx + vf y
= (!5.246 m/s)2 + (!5.129 m/s)2
r
vf = 7.3 m/s
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-16
#v &
fy
! = tan "1 %
(
%$ vfx ('
# 5.129 m/s &
= tan "1 %
(
$ 5.246 m/s '
! = 44°
Statement: The final velocity of the boat is 7.3 m/s [W 44° S].
!
!
12. Given: vf = 3.6 ! 102 km/h [N]; aav = 5.0 m/s 2 [W]; "t = 9.2 s
!
Required: vf
Analysis: Convert the final velocity to metres per second.
1h
1000 m
!
vf = 3.6 ! 102 km /h !
!
3600 s 1 km
!
2
vf = 1.0 ! 10 m/s [N]
Calculate the change in velocity from the acceleration and time interval:
!
!v
!
aav =
!t
! !
!v = aav !t
! ! !
!
Using components and !v = vf " vi , determine vi . Use east and north as positive.
Solution: The change in velocity is
! !
!v = aav !t
= (5.0 m/s 2 [W]) (9.2 s)
!
!v = 46 m/s [W]
!
The x-component of vi is
vix = vfx ! "vx
= 0 m/s ! (!46 m/s)
vix = 46 m/s
!
The y-component of vi is
viy = vf y ! "v y
= 100 m/s ! 0 m/s
viy = 100 m/s
Determine the initial velocity from its components.
2
2
r
vi = vix + viy
= (46 m/s)2 + (100 m/s)2
= 110.1 m/s
r
vi = 1.1! 102 m/s
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-17
#v &
iy
! = tan "1 %
(
%$ vix ('
# 100 m/s &
= tan "1 %
(
$ 46 m/s '
! = 65°
Statement: The initial velocity of the boat is 1.1! 102 m/s [E 65° N].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.4-18
Section 1.5: Projectile Motion
Mini Investigation: Analyzing the Range of a Projectile, page 38
Answers may vary. Sample answers:
A. There appears to be no relationship between the horizontal component of velocity and the
maximum height of a projectile.
B. The maximum height of a projectile is greatest when the launch angle is largest.
C. A projectile has maximum range when launched at an intermediate angle, around 45°.
D. The range of a projectile is the same whether launched at an angle or at the complement of
that angle. (Two angles are complementary if they add up to 90°; for example, 25° and 65° are
complementary angles, as are 11° and 79°.)
Tutorial 1 Practice, page 40
1. (a) Given: viy = 0 m/s; !d y = 76.5 cm = 0.765 m; g = 9.8 m/s 2
Required: Δt
Analysis: Set the table top as di = 0. Therefore Δ dy = –0.765 m. In the vertical direction, I know
the displacement, initial velocity, and acceleration. Use down as positive, so the displacement
1
will be negative. Use !d y = v1y !t " g!t 2 to determine Δt using the quadratic formula.
2
1
g!t 2 " v1y !t + !d y = 0
2
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
!t =
g
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
Solution: !t =
g
!t =
#1
&
0 ± 0 " 4 % (9.8 m/s 2 )( ("0.765 m)
$2
'
9.8 m/s 2
!t = ±0.3951 s (two extra digits carried)
!t = 0.40 s
Statement: The marble hits the floor after 0.40 s.
(b) Given: vx = 1.93 m/s; !t = 0.3951 s
Required: !d x
Analysis: Since I know the time of flight of the marble and its horizontal velocity. I can
determine its horizontal range using !d x = vx !t .
Solution: !d x = vx !t
= (1.93 m/s )(0.3951 s)
!d x = 0.76 m
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-1
Statement: The range of the marble is 0.76 m, or 76 cm.
(c) Given: vx = 1.93 m/s; viy = 0 m/s; !t = 0.3951 s; g = 9.8 m/s 2
!
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration: !v y = "g!t . Determine the vertical component
of the final velocity and then construct the final velocity vector.
Solution: Determine the vertical component of the final velocity.
!v y = "g!t
vf y = viy " g!t
= 0 m/s " (9.8 m/s 2 )(0.3951 s)
vf y = "3.872 m/s (two extra digits carried)
!
Since vfx = vx = 1.93 m/s , I can combine the components to determine the magnitude of vf .
!
vf = (vfx )2 + (vf y )2
= (1.93 m/s)2 + (!3.872 m/s)2
!
vf = 4.3 m/s
!
The angle below the horizontal axis of vf is
#v &
fy
! = tan %
(
%$ vf x ('
"1
# 3.872 m/s &
= tan "1 %
(
$ 1.93 m/s '
! = 64°
Statement: The final velocity of the marble is 4.3 m/s [64° below the horizontal].
2. Given: viy = 0 m/s; !d y = "0.83 m; !d x = 18.4 m; g = 9.8 m/s 2
Required: vx
Analysis: I know the horizontal displacement and want to determine the constant horizontal
speed. The appropriate formula is !d x = vx !t , but I do not know the time taken. Looking at the
vertical motion, use !d y = v1y !t "
1
g!t 2 " v1y !t + !d y = 0
2
1
g!t 2 to determine Δt using the quadratic formula.
2
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
!t =
g
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-2
Solution: Using the vertical motion,
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
!t =
g
!t =
#1
&
0 ± 0 " 4 % (9.8 m/s 2 )( ("0.83 m)
$2
'
9.8 m/s 2
!t = ±0.4116 s (two extra digits carried)
Using the horizontal motion,
!d x = vx !t
!d x
!t
18.4 m
=
0.4116 s
vx = 45 m/s
Statement: The initial horizontal speed of the ball is 45 m/s.
r
3. (a) Given: vi = 12 m/s [42° above the horizontal]; !d y = "9.5 m; g = 9.8 m/s 2
vx =
Required: Δt
Analysis: First, determine the components of the initial velocity. Then, use the vertical motion
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
1
and !d y = v1y !t " g!t 2 to solve for the time taken: !t =
g
2
Solution: The components of the initial velocity are
vix = (12 m/s)(cos 42°)
vix = 8.918 m/s (two extra digits carried)
viy = (12 m/s)(sin 42°)
viy = 8.030 m/s (two extra digits carried)
Using the vertical motion,
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
!t =
g
=
8.030 m/s ± ("8.030 m/s)2 " 4(4.9 m/s 2 )("9.5 m)
2(4.9 m/s 2 )
!t = 2.435 s or " 0.796 s (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-3
The time of flight cannot be a negative value.
!t = 2.4 s
Statement: The rock’s time of flight is 2.4 s.
(b) Given: vix = 8.918 m/s; !t = 2.435 s; g = 9.8 m/s 2
Required: width of the moat, !d x
Analysis: Since I know the time of flight of the rock and its horizontal velocity, I can determine
its horizontal range (the width of the moat) using !d x = vx !t .
Solution: !d x = vx !t
= (8.918 m/s)(2.435 s)
!d x = 22 m
Statement: The width of the moat is 22 m.
(c) Given: vix = 8.918 m/s; viy = 8.030 m/s; !t = 2.435 s; g = 9.8 m/s 2
!
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration: !v y = "g!t . Determine the vertical component
of the final velocity and then construct the final velocity vector.
Solution: Determine the vertical component of the final velocity.
!v y = "g!t
vf y = viy " g!t
= 8.030 m/s " (9.8 m/s 2 )(2.435 s)
vf y = "15.83 m/s (two extra digits carried)
r
Since vfx = vx = 8.918 m/s , I can combine the components to determine the magnitude of vf .
!
vf = (vfx )2 + (vf y )2
= (8.918 m/s)2 + (!15.83 m/s)2
= 18.17 m/s
!
vf = 18 m/s
r
The angle below the horizontal axis of vf is
#v &
fy
! = tan "1 %
(
%$ vf x ('
# 15.83 m/s &
= tan "1 %
(
$ 8.918 m/s '
! = 61°
Statement: The final velocity of the rock is 18 m/s [61° below the horizontal].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-4
!
4. (a) Given: vi = 4.3 m/s [42° below the horizontal]; !d y = –3.9 m + 1.4 m = "2.5 m;
g = 9.8 m/s 2
Required: Δt
Analysis: First determine the components of the initial velocity. Then use the vertical motion
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
1
and !d y = v1y !t " g!t 2 to solve for the time taken: !t =
g
2
Solution: The components of the initial velocity are
viy = !(4.3 m/s)(sin 42°)
vix = (4.3 m/s)(cos 42°)
vix = 3.196 m/s
viy = !2.877 m/s
Using the vertical motion,
"2.877 m/s ± (2.877 m/s)2 " 4(4.9 m/s 2 )(–2.5 m)
9.8 m/s 2
!t = 0.4787 s or " 1.066 s (two extra digits carried)
The time the ball is in the air cannot be a negative value.
!t = 0.48 s
Statement: The baseball’s time of flight is 0.48 s.
(b) Given: vix = 3.196 m/s; !t = 0.4787 s
!t =
Required: !d x
Analysis: Since I know the time of flight of the baseball and its horizontal velocity, I can
determine its horizontal range using !d x = vx !t .
Solution: !d x = vx !t
= (3.196 m/s)(0.4787 s)
!d x = 1.5 m
Statement: The horizontal distance from the window is 1.5 m.
(c) Given: vix = 3.196 m/s; viy = !2.877 m/s; "t = 0.4787 s; g = 9.8 m/s 2
!
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration: !v y = "g!t . Determine the vertical component
of the final velocity and then construct the final velocity vector.
Solution: Determine the vertical component of the final velocity.
!v y = "g!t
vf y = viy " g!t
= "2.877 m/s " (9.8 m/s 2 )(0.4787 s)
vf y = "7.568 m/s
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-5
r
Since vfx = vx = 3.196 m/s , I can combine the components to determine the magnitude of vf .
!
vf = (vfx )2 + (vf y )2
= (3.196 m/s)2 + (!7.568 m/s)2
!
vf = 8.2 m/s
Statement: The speed of the ball as you catch it is 8.2 m/s.
Tutorial 2 Practice, page 42
r
1. (a) Given: vi = 2.2 ! 102 m/s [45° above the horizontal]; diy = df y ; g = 9.8 m/s 2
Required: Δt
Analysis: Since the projectile (the marble) lands at the same height from which it was launched,
2v sin θ
the time taken is given by Δt = i
.
g
2v sin "
Solution: !t = i
g
2(220 m/s)(sin 45°)
=
9.8 m/s 2
!t = 32 s
Statement: The time of flight is 32 s.
!
(b) Given: vi = 2.2 ! 102 m/s [45° above the horizontal] ; g = 9.8 m/s2
Required: !d x
vi2 sin 2"
Analysis: Use the range formula, !d x =
g
vi2 sin 2"
Solution: !d x =
g
(220 m/s)2 (sin 90°)
9.8 m/s 2
!d x = 4.9 # 103 m
=
Statement: The horizontal range of the projectile is 4.9 ! 103 m , or 4.9 km.
!
(c) Given: vi = 2.2 ! 102 m/s [45° above the horizontal] ; g = 9.8 m/s2
Required: maximum height, d y
Analysis: The maximum height occurs when v y = 0 m/s . Use the acceleration formula
vf2y = viy2 ! 2g"d y to determine d y .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-6
Solution: Since viy = vi sin ! and vf y = 0 m/s , d y can be calculated.
vf2y = viy2 ! 2g"d y
"d y =
vf2y
2g
(vi sin # )2
"d y =
2g
((220 m/s)sin 45°)2
=
2(9.8 m/s 2 )
= 1235 m
"d y = 1.2 $ 103 m
Statement: The projectile’s maximum height is 1.2 × 103 m above the ground.
!
(d) Given: vi = 2.2 ! 102 m/s [45° above the horizontal]; diy = df y
r
Required: velocity on impact, vf
Analysis: The projectile’s flight is symmetric. Its final velocity is the same as the initial velocity
except that the direction of its vertical component is reversed.
Statement: The velocity of the marble when it hits the floor is 2.2 × 102 m/s [45º below the
horizontal].
r
2. (a) Given: vi = 14.5 m/s [35.0° above horizontal]; diy = df y ; g = 9.8 m/s 2
Required: maximum height, d y
Analysis: The maximum height occurs when v y = 0 m/s . Use the acceleration formula
vf2y = viy2 ! 2g"d y to determine d y .
vf2y = viy2 ! 2g"d y
"d y =
vf2y
2g
Solution: Since viy = vi sin ! and vf y = 0 m/s , d y can be calculated.
(vi sin " )2
!d y =
2g
((14.5 m/s)sin 35°)2
=
2(9.8 m/s 2 )
!d y = 3.5 m
Statement: The projectile’s maximum height is 3.5 m above the ground.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-7
!
(b) Given: vi = 14.5 m/s [35.0° above horizontal] ; g = 9.8 m/s2
Required: horizontal range, !d x
vi2 sin 2"
Analysis: Use the range formula !d x =
.
g
Solution: !d x =
vi2 sin 2"
g
(14.5 m/s)2 (sin 70.0°)
9.8 m/s 2
!d x = 2.0 # 101 m
=
Statement: The horizontal range of the projectile is 2.0 × 101 m.
!
(c) Given: vi = 14.5 m/s [35.0° above horizontal] ; g = 9.8 m/2
Required: time to maximum height, Δt
Analysis: By symmetry, the time to maximum height is half of the total time Δt =
Solution: !t =
2vi sin θ
.
g
vi sin "
g
(14.5 m/s)(sin 35.0°)
9.8 m/s 2
!t = 0.85 s
Statement: The time for the projectile to reach its maximum height is 0.85 s.
3. Solutions may vary. Sample answer:
2v sin θ
(a) The time of flight is given by Δt = i
. So Δt is proportional to vi . When vi doubles,
g
Δt doubles.
v 2 sin 2"
. So !d x is proportional to vi2 . When vi doubles, !d x
(b) The range is given by !d x = i
g
increases by a factor of four.
(vi sin " )2
(c) The maximum height is given by !d y =
. So !d y is proportional to vi2 . When vi
2g
doubles, !d y increases by a factor of four.
=
Section 1.5 Questions, page 43
1. Given: viy = 0 m/s; !d y = "1.5 m; !d x = 8.3 m; g = 9.8 m/s 2
Required: rock’s initial speed, vx
Analysis: I know the horizontal displacement and want to determine the constant horizontal
speed. The appropriate formula is !d x = vx !t , but I do not know the time taken.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-8
!d x = vx !t
!d x
!t
Looking at the vertical motion, determine Δt from
1
!d y = v1y !t " g!t 2
.
2
vx =
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
!t =
g
Solution: Using the vertical motion,
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
!t =
g
0 m/s ± (0 m/s)2 " 4(4.9 m/s 2 )("1.5 m)
9.8 m/s 2
!t = ±0.5533 s (two extra digits carried)
Using the horizontal motion,
!d
vx = x
!t
8.3 m
=
0.5533 s
vx = 15 m/s
Statement: The initial speed of the rock is 15 m/s.
r
2. (a) Given: vi = 1.1! 103 m/s [45° above the horizontal]; diy = df y ; g = 9.8 m/s 2
=
Required: Δt
Analysis: Since the projectile lands at the same height from which it was launched, the time
2v sin θ
.
taken is given by Δt = i
g
2v sin "
Solution: !t = i
g
=
2(1.1# 103 m / s )(sin 45°)
9.8 m /s 2
= 158.7 s
!t = 1.6 # 102 s
Statement: The object is in the air for 1.6 ! 102 s .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-9
!
(b) Given: vi = 1.1 ! 103 m/s [45° above the horizontal] ; g = 9.8 m/s2
Required: !d x
Analysis: Use the range formula !d x =
vi2 sin 2"
g
Solution: !d x =
=
=
vi2 sin 2"
.
g
(1.1# 103 m/s)2 (sin 90°)
9.8 m/s 2
(1.1# 103 m/ s )(1.1# 103 m / s )(sin 90°)
9.8 m / s 2
= 1.2 # 105 m
!d x = 1.2 # 102 km
Statement: The horizontal range of the projectile is 1.2 × 102 km.
r
(c) Given: vi = 1.1! 103 m/s [45° above the horizontal]; g = 9.8 m/s 2
Required: maximum height, d y
Analysis: Use the formula !d y =
Solution: !d y =
=
(vi sin " )2
2g
(vi sin " )2
2g
((1100 m/s)sin 45°)2
2(9.8 m/s 2 )
= 3.1# 104 m
!d y = 31 km
Statement: The projectile’s maximum height is 31 km above the ground.
r
3. (a) Given: vi = 6.0 m/s [32° below the horizontal]; !t = 3.4 s; g = 9.8 m/s 2
Required: !d y
Analysis: First determine the components of the initial velocity. Then use the vertical motion
1
and !d y = v1y !t " g!t 2 to solve for the vertical displacement.
2
Solution: The components of the initial velocity are
vix = (6.0 m/s)(cos 32°)
vix = 5.088 m/s (two extra digits carried)
viy = !(6.0 m/s)(sin 32°)
viy = !3.180 m/s (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-10
Using the vertical motion,
1
!d y = v1y !t " g!t 2
2
1
= ("3.180 m/s )(3.4 s ) " (9.8 m/ s 2 )(3.4 s )2
2
!d y = –67 m
Statement: The ball fell 67 m, so the window was 67 m above the ground.
(b) Given: vix = 5.088 m/s; viy = !3.180 m/s; "t = 3.4 s ; g = 9.8 m/s2
r
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration:
!v y = "g!t
vf y = viy " g!t
Determine the vertical component of the final velocity and then construct the final velocity
vector.
Solution: Determine the vertical component of the final velocity.
vf y = viy ! g"t
= !3.180 m/s ! (9.8 m/s 2 )(3.4 s )
vf y = !36.50 m/s (two extra digits carried)
r
Since vfx = vx = 5.088 m/s , I can combine the components to determine the magnitude of vf .
!
vf = (vfx )2 + (vf y )2
= (5.088 m/s)2 + (!36.50 m/s)2
!
vf = 37 m/s
r
The angle below the horizontal axis of vf is
#v &
fy
! = tan "1 %
(
%$ vfx ('
# 36.50 m/s &
= tan "1 %
(
$ 5.088 m/s '
! = 82°
Statement: The final velocity of the ball is 37 m/s [82° below the horizontal].
4. (a) Given: ball’s initial direction, ! = 53° above the horizontal; "d x = 25 m; "t = 2.1 s
r
Required: initial velocity, vi
Analysis: Draw a diagram of the situation. The horizontal component of velocity is constant.
Determine it from the time interval and the horizontal distance using !d x = vx !t . Then use the
cosine of the initial angle to determine the initial speed.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-11
Solution:
r
The x-component of vi is
!d x = vx !t
!d x
!t
25 m
=
2.1 s
vx = 11.91 m/s (two extra digits carried)
Determine the initial speed.
vix
cos! =
vi
vx =
vi =
vix
cos!
11.91 m/s
=
cos 53°
= 19.79 m/s (two extra digits carried)
vi = 2.0 " 101 m/s
Statement: The initial velocity of the soccer ball is 2.0 × 101 m/s [53° above the horizontal].
r
(b) Given: vi = 19.79 m/s [53° above the horizontal]; !d y = 7.2 m; g = 9.8 m/s 2
Required: horizontal range, !d x
Analysis: I know the horizontal component of the initial velocity and want the horizontal
distance. I can use !d x = vx !t once I know the time of flight. For this I need to look at the
vertical motion, viy = vi sin ! .
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
1
Use !d y = v1y !t " g!t 2 to determine !t =
.
g
2
Then solve the horizontal motion.
Solution: The vertical component of the initial velocity is
viy = vi sin !
= (19.79 m/s)( sin 53°)
viy = 15.80 m/s (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-12
#1 &
viy ± viy2 " 4 % g ( (!d y )
$2 '
!t =
g
15.80 m/s ± (–15.80 m/s)2 " 4(4.9 m/s 2 )(7.2 m)
9.8 m/s 2
!t = 2.675 s or 0.5495 s (two extra digits carried)
The ball lands on the building at about 2.7 s. The other time corresponds to the ball moving
through a height of 7.2 m on the way up.
The horizontal displacement of the ball is
!d x = vx !t
= (11.91 m/s )(2.675 s )
!d x = 32 m
Statement: The ball’s horizontal range is 32 m.
(c) Given: !d x = 25 m; vx = 11.91 m/s; viy = 15.80 m/s ; g = 9.8 m/s2
=
Required: ball’s clearance above wall based on !d y
Analysis: I have calculated the components of the initial velocity above. Use the x-component
and !d x = vx !t to determine the time for the ball to reach the building. Use the y-component and
1
g!t 2 to determine the ball’s height at that time. Then calculate how much above
2
7.2 m the ball is.
Solution: Calculate the time for the ball to reach the wall.
!d x = vx !t
!d y = v1y !t "
!t =
!d x
vx
25 m
11.91 m/s
!t = 2.101 s (two extra digits carried)
The vertical displacement of the ball at this time is
1
!d y = v1y !t " g!t 2
2
= (15.80 m/s)(2.101 s) " (4.9 m/s 2 )(2.101 s)2
=
= 33.19 m " 21.63 m
!d y = 11.56 m
Determine the distance by which the ball clears the wall.
!h = 11.56 m " 7.2 m
!h = 4.4 m
Statement: The ball clears the wall by 4.4 m.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-13
r
5. (a) Given: vi = 26 m/s [52° above horizontal]; vf y = 0 m/s; g = 3.7 m/s 2 ; g = 9.8 m/s 2
Required: maximum height, df y , based on !d y
Analysis: Determine the vertical component of the initial velocity using viy = vi sin ! .
Then use vf2y = viy2 ! 2g"d y .
Solution: The vertical component of the initial velocity is
viy = vi sin !
= (26 m/s)(sin 52°)
viy = 20.49 m/s (two extra digits carried)
The vertical displacement is
vf2y = viy2 ! 2g"d y
(0 m/s)2 = (20.49 m/s)2 ! 2(3.7 m/s 2 )"d y
"d y =
(20.49 m/s ) 2
2(3.7 m/ s 2 )
"d y = 57 m
Statement: The rock rises to a maximum height of 57 m.
!
(b) Given: vi = 26 m/s [52° above horizontal]; !d y = 12 m; g = 3.7 m/s 2
Required: time of flight, Δt
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
1
Analysis: Use !d y = v1y !t " g!t 2 to determine !t =
.
g
2
20.488 m/s ± ("20.488 m/s)2 " 4(1.85 m/s 2 )(12 m)
Solution: !t =
2(1.85 m/s 2 )
= 10.45 s or 0.621 s (two extra digits carried)
!t = 1.0 # 101 s
Statement: The rock strikes the hill after 1.0 ! 101 s .
!
(c) Given: vi = 26 m/s [52° above horizontal]; ! = 10.45 s
Required: horizontal range, !d x
Analysis: Calculate the horizontal component of the initial velocity using vix = vi cos! .
Using !d x = vx !t , determine the horizontal displacement.
r
Solution: The horizontal component of vi is
vix = vi cos!
= (26 m/s)(cos 52º )
vix = 16.01 m/s (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-14
The horizontal distance travelled is
!d x = vx !t
(
)
= 16.01 m/ s (10.45 s )
!d x = 1.7 " 102 m
Statement: The rock’s range is 1.7 ! 102 m .
r
6. Given: vi = 16 m/s [65° above horizontal]; angle of inclination of hill ! = 30°; g = 9.8 m/s 2
Required: distance up the hill where rock lands, l
Analysis: Draw a diagram of the situation. I know the initial velocity of the rock. From
1
!d x = vx !t and !d y = v1y !t " g!t 2 , I can determine where the rock is at any later time. The
2
"d y
rock will hit the hill when its displacement components are in the correct ratio: tan ! =
.
"d x
Solution:
Determine the x- and y-components of the initial velocity.
vix = vi cos!
= (16 m/s)( cos 65°)
vix = 6.762 m/s (two extra digits carried)
viy = vi cos!
= (16 m/s)( sin 65°)
viy = 14.50 m/s (two extra digits carried)
Write equations for the x- and y-components of the displacement after time Δt .
1
!d x = vx !t
!d y = v1y !t " g!t 2
2
!d x = (6.762 m/s)!t
!d y = (14.50 m/s)!t " (4.9 m/s 2 )!t 2
Use tan ! =
"d y
"d x
to build an equation for the time taken to reach the hill, Δt .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-15
tan ! =
"d y
"d x
(14.50 m/s)"t # (4.9 m/s 2 )"t 2
tan30° =
(6.762 m/s)"t
(0.5774)(6.762 m/s)"t = (14.50 m/s)"t # (4.9 m/s 2 )"t 2
(4.9 m/s 2 )"t 2 + (3.904 m/s)"t # (14.50 m/s)"t = 0
(4.9 m/s 2 )"t 2 # (10.60 m/s)"t = 0
[(4.9 m/s 2 )"t # (10.60 m/s)]"t = 0
One solution is Δt = 0 s, when the rock is thrown from the bottom of the hill. The other solution
is as follows:
(4.9 m/s 2 )!t " (10.60 m/s) = 0
10.60 m/s
4.9 m/s 2
!t = 2.163 s (two extra digits carried)
The x- and y-components of the displacement at this time are
!d x = (6.762 m/s)!t
!t =
= (6.762 m/s)(2.163 s)
!d x = 14.63 m (two extra digits carried)
!d y = (14.50 m/s)!t " (4.9 m/s 2 )!t 2
= (14.50 m/s )(2.163 s ) " (4.9 m/ s 2 )(2.163 s )2
= 31.37 m " 22.92 m
!d y = 8.45 m
The distance up the hill is
l = (14.63 m)2 + (8.45 m)2
l = 17 m
Statement: The rock lands 17 m up the hill.
7. Solutions may vary. Sample answer:
r
Given: vi = 45 m/s [35° above horizontal]; dix = 0 m; diy = 12 m; g = 9.8 m/s 2
Required: whether the snowball lands on a 25 m high, 35 m wide building, 150 m away
Analysis: One approach is to determine the horizontal displacement when the snowball is 25 m
up. If the snowball is between 150 m and 185 m horizontally from its launch position, then it is
over the building—and lands on the building.
Determine the components of the initial velocity, calculate the time to reach d y = 25 m , and then
determine d x at that time.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-16
Solution: The x- and y-components of the initial velocity are
vix = vi cos!
= (45 m/s)( cos 35°)
vix = 36.86 m/s (two extra digits carried)
viy = vi cos!
= (45 m/s)( sin 35°)
viy = 25.81 m/s (two extra digits carried)
The time to reach a height of 35 m starting at 12 m is
1
!d y = v1y !t " g!t 2
2
1
13 m = (25.81 m/s)!t " (9.8 m/s 2 )!t 2
2
2
2
(4.9 m/s )!t " (25.81 m/s)!t + 13 m = 0
Solve the quadratic equation for Δt .
(25.81 m/s) ± (25.81 m/s)2 " 4(4.9 m/s 2 )(13 m)
!t =
9.8 m/s 2
!t = 0.564 s or 4.703 s (two extra digits carried)
The first solution represents the snowball passing through the height of 25 m on its way up.
The second solution is the one where the snowball may be over the building. Determine the
horizontal displacement when Δt = 4.703 s .
!d x = (36.86 m/s)!t
= (36.86 m/s)(4.703 s)
!d x = 170 m
The horizontal displacement is indeed between 150 m and 185 m.
Statement: Yes, the snowball lands on top of building 2.
8. Solutions may vary. A written explanation and an algebraic solution are presented.
Written Explanation: All objects fall with the same acceleration regardless of mass. When a
projectile is fired at a target, its path is formed by the action of gravity together with its given
velocity. If gravity were to stop acting, the projectile would follow a straight line directly from
the launcher to the target. When gravity is acting, both the projectile and falling target fall at the
same speed, even though the projectile is travelling faster horizontally than the target. So at some
point along the path of the dropped target, the projectile will hit it because both the projectile and
the target are falling at the same speed. Given enough distance, they will both hit the floor at the
same time. If the projectile is fired with a greater velocity, the target will not fall as far before it
is hit by the projectile. In this case the projectile will follow a straighter path to the target. If the
projectile is fired more slowly, it will follow a more curved path and hit the target farther down
toward the ground. As long as the projectile launcher is aimed directly at the target and the
projectile has enough velocity to reach the target before it hits the ground, the projectile will hit
the target as it falls.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-17
Algebraic Solution:
!
!
!
Given: projectile’s motion: vi at ! above horizontal; di = 0 m; target: vi = 0 m/s;
diy = dix tan ! ; g = 9.8 m/s 2
Required: Show that the projectile hits the target.
Analysis: Write the equations for the position of the projectile after time Δt . Also do this for the
target. Then compare the equations to determine the time when the x-components of the two
objects coincide. Check whether the y-components coincide at the same time.
Solution:
projectile:
target:
!d x = vix !t
!d x = vix !t
!d x = vi cos" !t
!d x = 0 m
d x = vi cos" !t
(Equation 1)
d x = dix
(Equation 3)
1
!d y = v1y !t # g!t 2
2
1
d y = v1 sin " !t # g!t 2 (Equation 2)
2
1
!d y = v1y !t " g!t 2
2
1
!d y = (0 m/s)!t " g!t 2
2
1
d y = diy " g!t 2
(Equation 4)
2
Compare the horizontal positions of the projectile and the target using Equations 1 and 3.
vi cos! "t = dix
dix
vi cos
Compare the vertical positions of the projectile and the target using Equations 2 and 3, and also
the given fact diy = dix tan ! .
"t =
1
1
v1 sin ! "t # g"t 2 = diy # g"t 2
2
2
v1 sin ! "t = diy
"t =
=
diy
v1 sin !
dix tan !
v1 sin !
sin !
cos!
=
v1 sin !
dix
"t =
dix
vi cos!
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-18
!t =
=
diy
v1 sin "
dix tan "
v1 sin "
sin "
cos"
=
v1 sin "
dix
!t =
dix
vi cos"
The x-positions of the projectile and target are equal at the same time that the y-positions are
equal. In other words, the projectile hits the target.
Statement: The equations of motion describe motion at constant velocity, except for the
1
! g"t 2 term in the y-direction. Since this term is the same for both the projectile and target,
2
they will be equally affected by gravity. The projectile was aimed directly at the target, so it will
remain headed for the target as gravity pulls them both downward. So the projectile will hit the
target.
9. Solutions may vary. Sample answer:
!
Given: football: vi = 18 m/s [39° above horizontal]; dix = 0 m; diy = 22 m ;
r
player: v = 6.0 m/s [horizontally]; dix = 12 m; diy = 0 m; g = 9.8 m/s 2
Required: whether the player can catch the ball
Analysis: Draw a diagram for the situation. There is enough information to determine where and
1
when the football would hit the ground. Use !d y = v1y !t " g!t 2 to
2
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
determine !t =
.
g
Check whether or not the player has enough time to get to the correct position while running at
constant speed.
Solution:
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-19
Determine the vertical speed.
v1y = v1 sin !
= (18 m/s)(sin 39°)
v1y = 11.33 m/s (two extra digits carried)
Determine when the football hits the ground.
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
!t =
g
(11.33 m/s) ± (11.33 m/s)2 " 4(4.9 m/s 2 )("22 m)
=
9.8 m/s 2
!t = 3.570 s or " 1.258 s (two extra digits carried)
Time must be a positive quantity, so the time when the football hits the ground is 3.570 s.
!d x = vix !t
!d x = vi cos" !t
= (18 m/s)(cos 39°)(3.570 s)
!d x = 64 m
The football will hit the ground about 64 m from the cliff, about 3.6 s after being thrown.
At this time the player can make it to the position
!d x = vx !t
= (6.0 m/s)(3.570 s)
!d x = 21.42 m
d x = 12 m + 21.42 m
d x = 32 m
The player is about 32 m short of catching the football.
Statement: No, the player cannot run far enough to catch the ball.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.5-20
Section 1.6: Relative Motion
Tutorial 1 Practice, page 47–48
!
!
1. (a) Given: vSW = 2.8 m/s [forward]; vTS = 1.1 m/s [forward]
!
Required: vTS
!
!
!
Analysis: Use vTW = vTS + vSW , with forward as the positive direction.
!
!
!
Solution: vTW = vTS + vSW
= 1.1 m/s [forward] + 2.8 m/s [forward]
!
vTW = 3.9 m/s [forward]
Statement: The velocity of the teenagers with respect to the water is 3.9 m/s [forward].
!
!
(b) Given: vSW = 2.8 m/s [forward]; vTS = 1.1 m/s [backward]
!
Required: vTS
!
!
!
Analysis: Use vTW = vTS + vSW , with forward as the positive direction.
!
!
!
Solution: vTW = vTS + vSW
= 1.1 m/s [backward] + 2.8 m/s [forward]
= –1.1 m/s [forward] + 2.8 m/s [forward]
!
vTW = 1.7 m/s [forward]
Statement: The velocity of the teenagers with respect to the water is 1.7 m/s [forward].
!
!
2. Given: vPA = 235 km/h [N]; vAG = 65 km/h [E 45° N]
!
Required: vPG
Component Method:
Analysis: Use (vPG ) x = (vPA ) x + (vAG ) x and (vPG ) y = (vPA ) y + (vAG ) y , with east and north as
positive.
Solution: x-components:
(vPG ) x = (vPA ) x + (vAG ) x
= 0 km/h + (65 km/h)(cos 45°)
= 0 km/h + 45.96 km/h
(vPG ) y = 45.96 km/h (two extra digits carried)
y-components:
(vPG ) y = (vPA ) y + (vAG ) y
= 235 km/h + (65 km/h)(sin 45°)
= 235 km/h + 45.96 km/h
(vPG ) y = 281.0 km/h (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-1
!
Now use these components to determine vPG .
!
vPG =
(vPG ) x
2
+ (vPG ) y
2
= (45.96 km/h)2 + (281.0 km/h)2
= 284.7 km/h (two extra digits carried)
!
vPG = 280 km/h
# 281.0 km/h &
! 2 = tan "1 %
(
$ 45.96 km/h '
! 2 = 81°
Statement: The speed and direction of the plane with respect to the ground are 280 km/h
[E 81° N].
Geometry Method:
Analysis: I know two sides of the vector addition triangle, and the angle that lies in between.
Draw a diagram of the situation. Use the cosine and sine laws to determine the speed and
direction of the plane.
sin A sin B sinC
=
=
c 2 = a 2 + b2 ! 2abcosC
a
b
c
Solution:
From the vector addition diagram, ! 2 = 90° + 45° = 135° .
The cosine law gives
2
2
2
= vPA
+ vAG
! 2vPA vAG cos" 2
vPG
= (235 km/h)2 + (65 km/h)2 ! 2(235 km/h)(65 km/h)(cos 135°)
2
= 81 054 km/h
vPG
vPG = 284.7 km/h (two extra digits carried)
vPG = 280 km/h
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-2
The sine law gives
sin ! 3 sin ! 2
=
vAG
vPG
(65 km/h)(sin 135°)
284.7 km/h
! 3 = 9.29°
sin ! 3 =
! = 90° " ! 3
= 90º "9.29º
! = 81°
Statement: The velocity of the plane with respect to the ground is 280 km/h [E 81°N].
!
!
3. Given: vHA = 175 km/h [S]; vAG = 85 km/h [E]
!
Required: vHG
!
Analysis: The directions of the helicopter and the wind form a right angle with vHG as the
hypotenuse of a right-angled triangle. Use the Pythagorean theorem and the tangent ratio to
!
determine vHG .
!
Solution: Determine the magnitude of vHG .
2
2
vHG = vHA
+ vAG
= (175 km/h)2 + (85 km/h)2
vHG = 190 km/h
!
Determine the direction of vHG
!
# vHA &
"1
! = tan % ! (
$ vAG '
# 175 km/h &
= tan "1 %
(
$ 85 km/h '
! = 64°
Statement: The velocity of the helicopter with respect to the ground is 190 km/h [E 64°S].
!
!
4. Given: !d = 450 km [S]; !t = 3.0 h; vAG = 50.0 km/h [E]
!
Required: vPA
!
Analysis: The displacement of the plane is due south, so the direction of vPG is also south.
!
Determine the magnitude of vPG using the given displacement and time. Then draw the vector
!
!
!
addition diagram for vPG = vPA + vAG . Since this is a right-angled triangle, use the Pythagorean
!
theorem and the tangent ratio to determine vPA .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-3
Solution: The ground speed is
!d
vPG =
!t
450 km
=
3.0 h
vPG = 150 km/h
!
Determine the magnitude of vPA .
vPA = (vPG )2 + (vAG )2
= (150 km/h)2 + (50.0 km/h)2
vPA = 160 km/h
!
Determine the direction of vPA .
!
# vAG &
"1
! = tan % ! (
$ vPG '
# 50.0 km/h &
= tan "1 %
(
$ 150 km/h '
! = 18°
Statement: The plane should head [S 18° W] with an airspeed of 160 km/h.
!
!
5. (a) Given: vFE = 4.0 m/s [N]; vCF = 3.0 m/s [N]
!
Required: vCE
!
!
!
Analysis: Use vCE = vCF + vFE , with north as the positive direction.
!
!
!
Solution: vCE = vCF + vFE
= (+4.0 m/s) + (+3.0 m/s)
= +7.0 m/s
!
vCE = 7.0 m/s [N]
Statement: When the child is running north, the velocity of the child with respect to Earth is
7.0 m/s [N].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-4
!
!
(b) Given: vFE = 4.0 m/s [N]; vCF = 3.0 m/s [S]
!
Required: vCE
!
!
!
Analysis: Use vCE = vCF + vFE , with north as the positive direction.
!
!
!
Solution: vCE = vCF + vFE
= (+4.0 m/s) + (!3.0 m/s)
= +1.0 m/s
!
vCE = 1.0 m/s [N]
Statement: When the child is running south, the velocity of the child with respect to Earth
is 1.0 m/s [N].
!
!
(c) Given: vFE = 4.0 m/s [N]; vCF = 3.0 m/s [E]
!
Required: vCE
!
!
!
Analysis: Use vCE = vCF + vFE . This is a right-angled triangle, so use the Pythagorean theorem
and the tangent ratio.
!
Solution: Determine the magnitude of vCE .
vCE = (vCF )2 + (vFE )2
= (3.0 m/s)2 + (4.0 m/s)2
vCE = 5.0 m/s
!
Determine the direction of vCE .
!
# vFE &
"1
! = tan % ! (
$ vCF '
# 4.0 m/s &
= tan "1 %
(
$ 3.0 m/s '
! = 53°
Statement: When the child is running east, the child’s velocity with respect to Earth is 5.0 m/s
[E 53° N], or equivalently 5.0 m/s [N 37° E].
!
!
6. (a) Given: vPA = 3.5 ! 102 km/h [N 35° W]; vAG = 62 km/h [S]
!
Required: vPG
Analysis: Since one of the given velocities points due south, use the component method of
solution. Use (vPG ) x = (vPA ) x + (vAG ) x and (vPG ) y = (vPA ) y + (vAG ) y , with east and north as
positive.
Solution: x-components:
(vPG ) x = (vPA ) x + (vAG ) x
= (!350 km/h)(sin 35°) + 0 km/h
(vPG ) x = !200.8 km/h (two extra digits carried)
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Chapter 1: Kinematics
1.6-5
y-components:
(vPG ) y = (vPA ) y + (vAG ) y
= (350 km/h)(cos 35°) + (!62 km/h)
(vPG ) y = 224.7 km/h (two extra digits carried)
!
Now use these components to determine vPG .
!
vPG =
2
(vPG ) x + (vPG ) y
2
= (200.75 km/h)2 + (224.70 km/h)2
= 301.3 km/h (two extra digits carried)
!
vPG = 3.0 ! 102 km/h
# 200.8 km/h &
! = tan "1 %
(
$ 224.7 km/h '
! = 42°
Statement: The velocity of plane with respect to the ground is 3.0 ! 102 km/h [N 42° W].
!
(b) Given: vPG = 301.3 km/h [N 42º W]; !t = 1.2 h
!
Required: !d
r
r r
r
!d
; !d = vav !t .
Analysis: Since the plane moves at constant ground velocity, use vav =
!t
r r
Solution: !d = vPG !t
= (301.3 km/h [N 42° W])(1.2 h)
r
!d = 3.6 " 102 km/h [N 42° W]
Statement: The plane’s displacement after 1.2 h is 3.6 ! 102 km/h [N 42° W] .
!
!
7. (a) Given: vPW = 0.70 m/s [N]; vWE = 0.40 m/s [E]
!
Required: vPE
r
Analysis: The directions of the current and the swimmer form a right angle with vPE as the
hypotenuse of a right-angled triangle. Use the Pythagorean theorem and the tangent ratio to
!
determine vPE .
!
Solution: Determine the magnitude of vPE .
vPE = (vPW )2 + (vWE )2
= (0.70 m/s)2 + (0.40 m/s)2
vPE = 0.81 m/s
!
Determine the direction of vPE .
!
# vWE &
"1
! = tan % ! (
$ vPW '
# 0.40 m/s &
= tan "1 %
(
$ 0.70 m/s '
! = 30°
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-6
Statement: The velocity of the swimmer with respect to Earth is 0.81 m/s [N 30° E].
!
(b) Given: !d = 84 m [N]
Required: time to cross river, !t
r
r
!d
!d
!
!
Analysis: The component of vPE pointing north is vPW . Use vav =
.
; !t =
!t
vav
Solution: !t =
!d
vPW
84 m
0.70 m/s
!t = 1.2 " 102 s
Statement: It takes the swimmer 1.2 × 102 s to cross the river.
!
(c) Given: vWE = 0.40 m/s [E]; !t = 120 s
Required: distance downstream, !d
r
r r
r
!d
!
!
; !d = vav !t .
Analysis: The component of vPE pointing east is vWE . Use vav =
!t
Solution: !d = vWE !t
=
= (0.40 m/s )(120 s)
!d = 48 m
Statement: The swimmer lands 48 m downstream.
r
r
(d) Given: vPW = 0.70 m/s; vWE = 0.40 m/s [E]; vPE = ? [N]
!
Required: direction of vPW
!
Analysis: The vectors form a right-angled triangle with vPW as the hypotenuse. is a right-angled
r
!
triangle. Use the sine ratio to determine the direction of vPW with respect to vPE [N] :
sin ! =
opposite side
hypoteuse
# opposite side &
! = sin "1 %
$ hypoteuse ('
!
Solution: Determine the direction of vPW .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-7
!
# vWE &
! = sin % ! (
$ vPW '
"1
# 0.40 m/s &
= sin "1 %
(
$ 0.70 m/s '
! = 35°
Statement: The swimmer should head [N 35° W] to land directly north of her starting point.
!
!
!
!
8. Given: vC W = vC W ; vC E = 1.2 m/s [upstream]; vC E = 2.9 m/s [downstream]
1
2
1
2
Required: vWE
!
!
!
!
!
!
Analysis: I know the relative velocity equations, vC E = vC W + vWE and vC E = vC W + vWE .
1
1
2
2
!
!
!
Switching to a simpler notation, v = vC W = vC W and w = vWE . I will rewrite the relative velocity
1
2
equations, using downstream as the positive direction. Then, I can solve for the required speed.
Solution: The relative velocity equations are
!
!
!
vC E = vC W + vWE
1
1
!1.2 m/s = (!v) + w
(Equation 1)
!
!
!
vC E = vC W + vWE
2
2
+2.9 m/s = v + w
(Equation 2)
Adding Equations 1 and 2,
(!1.2 m/s) + (+2.9 m/s) = (!v) + w + v + w
1.7 m/s = 2w
w = 0.85 m/s
Statement: The speed of the water relative to Earth is 0.85 m/s.
!
(b) Given: vC E = 2.9 m/s [downstream]; vWE = w = 0.85 m/s
2
!
!
Required: v = vC W = vC W
1
2
Analysis: Substitute the speed of the water from part (a) into either Equation 1 or Equation 2.
Solution: Using Equation 2,
(+2.9 m/s) = v + w
v = (+2.9 m/s) ! w
= (+2.9 m/s) ! (0.85 m/s)
v = 2.0 m/s
Statement: The canoeists paddle at 2.0 m/s with respect to the water.
!
!
!
9. (a) Given: vPA = 630 km/h [N]; vAG = 35 km/h [S]; !d = 750 km [N]
Required: !t
!
!
!
Analysis: Use vPG = vPA + vAG , with north as the positive direction, to calculate the ground
!
!d
!d
!
velocity of the plane. Then use vav =
to determine the time !t =
.
!t
vav
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-8
Solution: The ground velocity is
!
!
!
vPG = vPA + vAG
= (+630 km/h) + (!35 km/h)
= +595 km/h
!
vPG = 595 km/h [N] (one extra digit carried)
The required time, !t , is
!d
!t =
vPG
=
750 km
595 km /h
!t = 1.3 h
Statement: The flight time is 1.3 h when the wind is blowing south.
!
!
!
(b) Given: vPA = 630 km/h [N]; vAG = 35 km/h [N]; !d = 750 km [N]
Required: !t
!
!
!
Analysis: Use vPG = vPA + vAG , with north as the positive direction, to calculate the ground
!
!d
!
to determine the time !t .
velocity of the plane. Then use vav =
!t
Solution: The ground velocity is
!
!
!
vPG = vPA + vAG
= (+630 km/h) + (+35 km/h)
= +665 km/h
!
vPG = 665 km/h [N] (one extra digit carried)
The required time !t is
!d
vav =
!t
!d
!t =
vPG
=
750 km
665 km /h
!t = 1.1 h
Statement: The flight time is 1.1 h when there is a tail wind. This time is shorter than when
there is an opposing wind because the plane moves more quickly with respect to the ground.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-9
!
!
(c) Given: vPA = 630 km/h; vAG = 35 km/h [E]; !d = 750 km [N]
!
Required: direction of vPA , !t
Analysis: The pilot needs to head somewhat west of north to compensate for the wind that heads
!
!
!
east. Sketch the relative velocities, vPG = vPA + vAG . Since the vector addition triangle is a rightangled triangle, use the Pythagorean theorem and the sine ratio to determine the direction of the
air velocity and the magnitude of the ground velocity. Then determine the flight time using
r
r
!d
!d
.
; !t =
vav =
!t
vav
Solution:
!
Determine the direction of vPA .
!
# vAG &
"1
! = sin % ! (
$ vPA '
# 35 km/h &
= sin "1 %
(
$ 630 km/h '
! = 3.2°
!
Determine the magnitude of vPG .
(vPA )2 = (vPG )2 + (vAG )2
vPG = (vPA )2 ! (vAG )2
= (630 km/h)2 ! (35 km/h)2
= 629.0 km/h (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-10
!t =
=
!d
vPG
750 km
629.0 km /h
!t = 1.2 h
Statement: The pilot’s heading needs to be [N 3.2° W]. The new flight time is 1.2 h.
Section 1.6 Questions, page 49
!
!
!
1. (a) Given: vPW = 1.2 m/s; vWE = 0.50 m/s [E]; !d1 = 1.0 km [W]; !d2 = 1.0 km [E]
Required: !t = !t1 + !t2
Analysis: Look first at the upstream motion. Determine the speed !of the person with respect to
!d
!
!
!
!
Earth using vPE = vPW + vWE . Then, rearrange the equation vav =
to determine the time
!t
!d
. Repeat this procedure for the downstream motion. Finally, use
required; !t =
vav
!t = !t1 + !t2 . Throughout, use east as the positive direction.
Solution:
upstream motion:
downstream motion:
r
r
r
!
!
!
vPE = vPW + vWE
vPE = vPW + vWE
= (!1.2 m/s) + (+0.50 m/s)
= (+1.2 m/s) + (+0.50 m/s)
= !0.70 m/s
!d
!t1 =
vPW
= +1.70 m/s
!d
!t2 =
vPW
1000 m
0.70 m/s
!t1 = 1429 s (two extra digits carried)
The total time for the swim is
!t = !t1 + !t2
1000 m
1.70 m/s
!t2 = 588.2 s (two extra digits carried)
=
=
= 1429 s + 588.2 s
= 2017 s "
1 min
60 s
!t = 34 min
Statement: The swim upstream and back takes 34 min.
(b) Answers may vary. Sample answer: No, the time will not change. The total time downstream
and back will also be 34 min. The first leg of the swim will be fast (588 s) and the second leg
slow (1429 s), but the whole swim takes the same time.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-11
(c) Given: v = 1.2 m/s; !d = 2.0 km
Required: !t
!d
!d
Analysis: vav =
; !t =
!t
vav
!d
v
2000 m
=
1.2 m/s
Solution: !t =
= 1667 s "
1 min
(two extra digits carried)
60 s
!t = 28 min
Statement: The total swim would take 28 min in still water. This time is less than for the swims
in parts (a) and (b). Swimming against the current is a great disadvantage that is not compensated
for fully by swimming with the current for the same distance. The trip takes longer when there is
a current because the current slows the swimmer when swimming against it.
!
!
2. (a) Given: vPA = 200 m/s [W]; vAG = 60 m/s [N]
!
Required: vCE
!
!
!
Analysis: vPG = vPA + vAG . This is a right-angled triangle, so use the Pythagorean theorem and
the tangent ratio.
!
Solution: Determine the magnitude of vPG .
(vPG )2 = (vPA )2 + (vAG )2
= (200 m/s)2 + (60 m/s)2
(vPG )2 = 43 600 m 2 /s 2
vPG = 209 m/s (two extra digits carried)
vPG = 200 m/s
!
Determine the direction of vPG .
!
# vAG &
"1
! = tan % ! (
$ vPG '
# 60 m/s &
= tan "1 %
(
$ 209 m/s '
! = 20°
Statement: The ground velocity of the plane is 200 m/s [W 20° N].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-12
!
!
!
(b) Given: vPA = 200 m/s [W]; vAG = 60 m/s [E]; !d = 300 km [W]
Required: !t
!
!
!
Analysis: First, use vPG = vPA + vAG to determine the ground velocity. Then rearrange the
!
!d
!d
!
equation vav =
to calculate the flight time; !t =
.
!t
vav
Solution: The ground velocity is
r
r
r
vPG = vPA + vAG
= 200 m/s [W] + 60 m/s [E]
= 200 m/s [W] + (–60 m/s [W])
r
vPG = 140 m/s [W] (one extra digit carried)
!t =
!d
vPG
300 km
1000 m
"
140 m/s
1 km
1 min
= 2143 s "
60 s
!t = 36 min
Statement: The flight time is 36 min.
!
!
3. (a) Given: vHA = 62 m/s [N]; vAE = 18 m/s [N]
!
Required: vHE
!
!
!
Analysis: vHE = vHA + vAE
r
r
r
Solution: vHE = vHA + vAE
=
= 62 m/s [N] + 18 m/s [N]
r
vHE = 8.0 ! 101 m/s [N]
Statement: The ground velocity of the helicopter is 8.0 × 101 m/s [N].
!
!
(b) Given: vHA = 62 m/s [N]; vAE = 18 m/s [S]
!
Required: vHE
!
!
!
Analysis: vHE = vHA + vAE
!
!
!
Solution: vHE = vHA + vAE
= 62 m/s [N] + 18 m/s [S]
= 62 m/s [N] + (–18 m/s [N])
!
vHE = 44 m/s [N]
Statement: The ground velocity of the helicopter is 44 m/s [N].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-13
!
!
(c) Given: vHA = 62 m/s [N]; vAE = 18 m/s [W]
!
Required: vHE
!
!
!
!
Analysis: The vectors vHE = vHA + vAE form a right-angled triangle with vHE as the hypotenuse.
!
Use the Pythagorean theorem and the tangent ratios to determine vHE .
Solution: vHE = (vHA )2 + (vAE )2
= (62 m/s)2 + (18 m/s)2
vHE = 65 m/s
!
Determine the direction of vHE .
!
# vAE &
"1
! = tan % ! (
$ vHA '
# 18 m/s &
= tan "1 %
(
$ 62 m/s '
! = 16°
Statement: The velocity of the helicopter with respect to Earth is 65 m/s [N 16° W].
!
!
(d) Given: vHA = 62 m/s [N]; vAE = 18 m/s [N 42° W]
!
Required: vHE
!
Analysis: Draw the vector addition diagram for the situation. Use components to determine vHE .
Use east and north as positive.
Solution:
x-components:
(vHE ) x = (vHA ) x + (vAE ) x
= 0 m/s + (!(18 m/s)sin 42°)
= 0 m/s + (!12.04 m/s)
(vHE ) x = !12.04 m/s (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-14
y-components:
(vHE ) y = (vHA ) y + (vAE ) y
= 62 m/s + (18 m/s)(cos 42°)
= 62 m/s + 13.38 m/s
(vHE ) y = 75.38 m/s (two extra digits carried)
!
Determine the magnitude of vHE .
vHE =
2
(vHE ) x + (vHE ) y
2
= (12.04 m/s)2 + (75.38 m/s)2
vHE = 76 m/s
!
Determine the direction of vHE .
# ( v! ) &
"1
! = tan % !HE x (
%$ ( vHE ) y ('
# 12.04 m/s &
= tan "1 %
(
$ 75.38 m/s '
! = 9.1°
Statement: The velocity of the helicopter with respect to the ground is 76 m/s [N 9.1° W].
!
4. (a) Given: vPW = 0.65 m/s [S]; !d x = 88 m [W]; !d y = 130 m [S]; river flows [W]
!
Required: vWE
Analysis: Sketch the river and the displacement vectors.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-15
!
!
!
Also draw vPE = vPW + vWE .
!
!
!
Vectors vPW and vWE are perpendicular components of vPE .
I know the river flows west and that the swimmer drifts 88 m west during her crossing. I can
determine the time !t of the crossing because she swims heading straight across the 130 m at
!d y
!d
0.65 m/s. Use vav =
to determine the crossing time and the vav = x to determine the speed
!t
!t
of the current. Use south and west as the positive directions.
Solution:
Crossing the river:
Drifting downstream:
!d y
!d
vav = x
vav =
!t
!t
88 m
!d y
vWE =
!t =
200 s
vPW
vWE = 0.44 m/s
130 m
=
0.65 m/s
!t = 200 s
Statement: The water moves at 0.44 m/s [W].
(b) Given: !d x = 88 m [W]; !d y = 130 m [S]
!
Required: vPE
Analysis: The displacement triangle and the velocity triangle are similar, right-angled triangles.
!
Use the displacement triangle to determine the angle θ, and the velocity triangle to calculate vPE .
Solution: In the displacement triangle:
!
$ #d '
x
"1
! = tan & ! )
&% #d y )(
$ 88 m '
= tan "1 &
% 130 m )(
! = 34°
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-16
In the velocity triangle:
(vPE )2 = (vPW )2 + (vWE )2
= (0.65 m/s)2 + (0.44 m/s)2
vPE = 0.78 m/s
Statement: The velocity of the swimmer with respect to Earth is 0.78 m/s [S 34° W].
!
!
(c) Given: vPW = 0.65 m/s ; vWE = 0.44 m/s [W]; vPE = ? [S]
!
Required: direction of vPW
!
!
!
Analysis: Use vPE = vPW + vWE to draw the relative velocities. This is a right-angled triangle. Use
!
the sine ratio to determine the direction of vPW .
r
r
# vWE &
vWE
"1
sin ! = r ; ! = sin % r (
vPW
$ vPW '
!
# vWE &
"1
Solution: ! = sin % ! (
$ vPW '
# 0.44 m/s &
= sin "1 %
(
$ 0.65 m/s '
! = 43°
Statement: The swimmer should head [S 43° E].
!
!
5. Given: !d = 1.4 " 103 km [S 43° E]; !t = 3.4 h; vAG = 55 km/h [S]
!
Required: vPA
!
!
!
Analysis: Draw the relative velocities using vPG = vPA + vAG . Solve the triangle using the cosine
and sine laws:
c 2 = a 2 + b2 ! 2abcosC
sin A sin B sinC
=
=
b
c
a
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-17
Solution:
r
r
!d
vPG =
!t
1.4 " 103 km [S 43° E]
=
3.4 h
= 411.8 km/h [S 43° E] (two extra digits carried)
(vPA )2 = (vPG )2 + (vAG )2 ! 2(vPG )(vAG )cos" 2
= (vPG )2 + (vAG )2 ! 2(vPG )(vAG )cos" 2
(vPA )2 = (411.8 km/h)2 + (55 km/h)2 ! 2(411.8 km/h)(55 km/h)(cos 43°)
vPA = 373.4 km/h (two extra digits carried)
vPA = 370 km/h
sin ! 3 sin ! 2
=
vPG
vPA
sin ! 3 =
=
vPG sin ! 2
vPA
(411.8 km/h )(sin 43°)
373.4 km/h
! 3 = 131.2°
! = 180º " ! 3
= 180º "131.2°
! = 49°
Statement: The plane must maintain an air velocity of 3.7 ! 102 km/h [S 49° E].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-18
!
!
6. (a) Given: !d = 220 km [N]; vAG = 42 km/h [N 36° E]; vPA = 230 km/h [?]
!
Required: direction of vPA
Analysis: The direction of the ground velocity (the track) of the plane is [N] and I know the
!
!
!
wind direction. So I can draw vPG = vPA + vAG .
The direction of the air speed (the heading), angle ! in the velocity triangle, can be determined
using the sine law.
sin ! sin 36°
=
Solution:
vPA
vAG
sin ! =
=
vAG sin 36°
vPA
(42 km/h)(sin 36°)
(230 km/h)
! = 6.167° (two extra digits carried)
! = 6.2°
Statement: The heading of the plane should be [N 6.2° W].
!
!
(b) Given: !d = 220 km [N]; vAG = 42 km/h [N 36° E]; vPA = 230 km/h; " = 6.167°
Required: !t
Analysis: Use the vector triangle and the sine law to calculate ground speed, vPG .
!d
!d
.
; !t =
!t
vPG
Solution: Determine the third angle in the vector triangle, ! .
! + " + 36° = 180°
The calculate !t using vPG =
! = 180° # " # 36°
! = 180° # 6.167° # 36°
! = 137.8° (two extra digits carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-19
sin ! sin 36°
=
vPA
vPG
vPA sin !
sin 36°
(230 km/h)(sin 137.8°)
=
sin 36°
= 262.6 km/h (two extra digits carried)
vPG =
vPG
!t =
=
!d
vPG
220 km
262.6 km /h
!t = 0.84 h
Statement: The trip will take 0.84 h.
!
!
7. (a) Given: vPA = 250 m/s [W]; vAG = 50.0 m/s [E]
Required: vPG
!
!
!
Analysis: Use vPG = vPA + vAG to determine vPG .
!
!
!
Solution: vPG = vPA + vAG
= 250 m/s [W] + 50.0 m/s [E]
= 250 m/s [W] + (–50.0 m/s [W])
!
vPG = 2.0 ! 102 m/s [W]
Statement: The ground speed of the plane on its way west is 2.0 ! 102 m/s .
!
!
(b) Given: vPA = 250 m/s [E]; vAG = 50.0 m/s [E]
Required: vPG
!
!
!
Analysis: vPG = vPA + vAG
!
!
!
Solution: vPG = vPA + vAG
= 250 m/s [E] + 50.0 m/s [E]
!
vPG = 3.0 ! 102 m/s [E]
Statement: The ground speed of the plane on its way east is 3.0 ! 102 m/s .
!
!
8. (a) Given: vPS = 2.0 m/s [up]; vSW = 3.2 m/s [E]
!
Required: vPW
!
!
!
!
Analysis: The relative velocity triangle, vPW = vPS + vSW , is right-angled with vPW as the
hypotenuse. Use the Pythagorean theorem and the tangent ratio to determine the magnitude and
!
direction of vPW .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-20
Solution: (vPW )2 = (vPS )2 + (vSW )2
vPW = (vPS )2 + (vSW )2
= (2.0 m/s)2 + (3.2 m/s)2
vPW = 3.8 m/s
#v &
! = tan "1 % PS (
$ vSW '
# 2.0 m/s &
= tan "1 %
(
$ 3.2 m/s '
! = 32°
Statement: The people who take the elevator move at 3.8 m/s [E 32° up] relative to the water.
!
!
(b) Given: vPS = 2.0 m/s [E 38° up]; vSW = 3.2 m/s [E]
!
Required: vPW
!
!
!
Analysis: Sketch the relative velocity triangle, vPW = vPS + vSW . Use components to determine
!
vPW , with east and up as positive.
Solution:
x-components:
(vPW ) x = (vPS ) x + (vSW ) x
= +(2.0 m/s)cos38° + (+3.2 m/s)
= 1.576 m/s + 3.2 m/s
(vPW ) x = 4.776 m/s (two extra digits carried)
y-components:
(vPW ) y = (vPS ) y + (vSW ) y
= (2.0 m/s)sin 38° + 0 m/s
= 1.231 m/s + 0 m/s
(vPW ) y = 1.231 m/s (two extra digits carried)
!
Now use these components to determine vPW .
!
vPW =
2
(vPW ) x + (vPW ) y
2
= (4.776 m/s)2 + (1.231 m/s)2
!
vPG = 4.9 m/s
Copyright © 2012 Nelson Education Ltd.
# 1.231 m/s &
! = tan "1 %
(
$ 4.776 m/s '
! = 14°
Chapter 1: Kinematics
1.6-21
Statement: The people who take the stairs move at 4.9 m/s [E 14° up] relative to the water.
!
!
9. (a) Given: vCE = 60.0 km/h [E]; vRC = ? [down 70.0° W]
!
Required: vRC
Analysis: Make a sketch of the car with the rain.
!
!
!
Also draw vRE = vRC + vCE .
Convert kilometres per hour to metres per second.
1h
km 1000 m
!
!
= 16.667 m/s (two extra digits carried)
3600 s
h
1 km
Solve for vRC using the sine ratio.
60
Solution: sin70° =
vCE
vRC
vCE
sin70.0°
16.667 m/s
=
sin70.0°
= 17.737 m/s (two extra digits carried)
vRC =
vRC = 17.7 m/s
Statement: The rain moves at 17.7 m/s [down 70.0° W] with respect to the car.
!
(b) Given: vRC = 17.737 m/s [down 70.0° W]
!
Required: vRE
Analysis: Look at the triangles in part (a). Solve for vRE using the cosine ratio.
Solution: cos70° =
vRE
vRC
vRE = vRC cos 70.0°
= (17.737 m/s)(cos 70.0°)
vRE = 6.07 m/s
Statement: The velocity of the rain with respect to Earth is 6.07 m/s [down].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-22
!
!
10. (a) Given: vPG = ? [N 30.0° W]; vAG = 48 km/h [W]; vPA = 260 km/h
!
Required: direction of vPA
Analysis: The direction of the ground velocity (the track) of the plane is [N 30.0° W], and I
!
!
!
!
know the wind direction, so I can draw vPG = vPA + vAG . The angle opposite vPA in the relative
velocity triangle is 60.0º (since [N 30.0° W] is equivalent to [W 60.0° N] ).
Calculate the direction of the air speed (the heading) from the angles ! 2 and ! 3 in the scalene
relative-velocity triangle using the sine law.
Solution: Using the sine law,
sin ! 2 sin 60.0°
=
vPA
vAG
sin ! 2 =
vAG sin 60.0°
vPA
(48 km/h)(sin60.0°)
260 km/h
! 2 = 9.2°
=
! 3 = 180° " 60.0° " ! 2
= 180° " 60.0° " 9.2°
! 3 = 110.8°
! = 180° " ! 3
= 180° " 110.8°
! = 69°
The heading of the plane should be [W 69° N], which is equivalent to [N 21° W].
Statement: The heading of the plane should be [N 21° W].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-23
!
(b) Given: vPG = ? [W 60.0° N]; vPA = 260 km/h; ! 3 = 110.8º
Required: vPG
Analysis: Use the relative-velocity triangle from part (a) and the sine law to determine vPG .
Solution:
sin ! 3 sin 60.0°
=
vPA
vPG
vPA sin ! 3
sin 60.0°
(260 km/h)(sin 110.8°)
=
sin 60.0°
vPG = 280 km/h
Statement: The ground speed of the plane is 280 km/h.
vPG =
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1.6-24
Chapter 1 Self-Quiz, page 53
1. (b)
2. (d)
3. (b)
4. (c)
5. (a)
6. (d)
7. (d)
8. False. The instantaneous velocity at a particular time is the slope of the position–time curve at
that time.
9. True
10. False. The addition of two displacement vectors does not depend on the order in which they
are added.
11. False. If the velocity vector of an object changes only in direction, the average acceleration is
not zero.
12. True
13. False. A stone projected horizontally from a cliff will reach the ground at the same time as a
stone dropped vertically down from the same cliff.
14. True
r
!
15. False. If vAB = 18.3 m/s [S] , then v AB = –18.3 m/s [N] .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-1
Chapter 1 Review, pages 54–59
Knowledge
1. (a)
2. (b)
3. (a)
4. (a)
5. (c)
6. (b)
7. True
8. False. The average speed is the slope of the secant drawn between the initial and final times of
interest on the position–time curve for the motion.
9. True
10. False. When a moving objects starts to slow down on a straight track, the average
acceleration of the object at any time interval after it starts slowing down is negative.
11. True
12. False. The magnitude of the average velocity of an object is always less than or equal to the
average speed.
13. True
14. False. Two students running toward each other with the same speed have opposite velocity
vectors relative to each other.
15. (a) Answers may vary. Sample answer: No, since the car is moving on a straight track the
magnitude of the velocity is the speed. If the velocity is constant then the speed is constant.
(b) Answers may vary. Sample answer: Yes, the ratio of the magnitude of the velocity and the
speed is equal to one. On a straight track, the velocity may be positive or negative, but its
magnitude is the speed.
16. Answers may vary. Sample answer: Yes, it is possible for an object to have varying velocity
and constant speed. An example is a car going around a corner at constant speed.
17. Answers may vary. Sample answer: Yes, two displacement vectors of the same length can
have a vector sum of zero. Consider a displacement of 10 m [N] followed by a displacement of
10 m [S]. Their vector sum is zero.
18. Answers may vary. Sample answer: It is not possible for a sprinting football player to stop
instantly because this would require a change of velocity (in this case, a slowing down) in a time
interval of !t = 0 . Nothing can occur in an infinitesimally short length of time.
19. When a skier jumps off a ramp and air resistance is not negligible, both the range and the
landing speed are reduced. Air resistance causes a slowing down in the horizontal direction,
shortening the range. It also reduces the vertical acceleration, so the vertical component of
velocity is lower throughout the jump. Together, the two reduced velocity components result in a
lower landing speed.
20. Answers may vary. Sample answer: Imagine that you are skiing down a slope and your hat
flies off. You see the hat caught in the wind, moving backward—this is the motion of the hat
relative to you. A snowboarder at rest on the ski slope sees the hat following behind you and
eventually slowing down—this is the motion of the hat relative to the snowboarder.
21. Answers may vary. Sample answer: When two objects are moving at different velocities,
each appears to be in motion as seen from the other. This motion is called relative motion
because the details of the motion are relative, depending on the point of view.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-2
Understanding
22. Answers may vary. Sample answer: An object can have acceleration because its speed
changes or because its direction changes. For example, an object moving around in a circle at a
constant speed has acceleration due to its change in direction.
23. Answers may vary. Sample answer:
(a) A car moving to the right and speeding up has both positive velocity and acceleration.
(b) A car moving to the right and slowing down has positive velocity and negative acceleration.
(c) A car moving to the right at constant speed has positive velocity and zero acceleration.
24. Answers may vary. Sample answer: A position–time graph for an object in uniform motion is
a straight line. The slope of the graph represents the object’s speed. Positive slope is forward
motion, and negative slope is backward motion. The intercept on the position axis gives the
initial position of the object. The velocity–time graph for the same uniform motion is a
horizontal line, intersecting the velocity axis at the constant velocity of the motion.
25. Answers may vary. Sample answer: The odometer records distance travelled since the car
was new or the odometer was reset. If you take odometer readings at two times, you can
calculate the average speed over the time interval. The speedometer records the instantaneous
speed of the car. If you take speedometer readings at two times, you can calculate the average
acceleration over the time interval.
26. Answers may vary. Sample answer: A ball tossed into the air and caught at the same height
from which it was thrown has symmetric upward and downward motions. Its average velocity is
zero. The only time during the flight that its instantaneous velocity is zero is when the ball is at
its maximum height.
27. For three displacement vectors to have a vector sum of zero, the vector resulting from the
addition of two of the vectors must be the negative of the third vector, that is, it must have the
same magnitude as the third vector but point in the opposite direction.
For a sum of 0, the endpoint of the third vector will be the same as starting point of the first
vector.
28. Answers may vary. Sample answer: An example in which an object moves in two
dimensions but has acceleration in one dimension is a projectile that experiences negligible air
resistance. The projectile has motion in both the horizontal and vertical direction but its
acceleration is only vertical.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-3
29. Answers may vary. Sample answer: A ball is tossed vertically upward from a roof and lands
again on the roof. Displacement is a comparison of two positions. An observer on the roof and an
observer on the ground may record the position of the ball differently but they will calculate the
same displacement. Because velocity is based on displacement, the two observers see identical
velocities.
30. Answers may vary. Sample answer: When you predict the direction of motion of an object
you are predicting the direction of the displacement vector of the object over a very short time
interval from now. Velocity is based on displacement and time taken, so the velocity vector is
parallel to the displacement vector and shows the desired direction. Acceleration is based on a
change in velocity. The acceleration vector is not always parallel to the velocity vector. In fact,
the acceleration vector predicts which way the motion of an object will curve.
31. Answers may vary. Sample answer: In the long jump, an athlete launches himself into
projectile motion. His horizontal speed running up to the jumping pad and the strength and
direction of his push off the pad will affect the initial speed and launch angle. A high initial
speed and a launch angle close to but below 45° will maximize the range, or horizontal distance,
of the jump. Also, any posture or clothing the athlete can use to minimize air resistance will help
lengthen the jump.
32. Answers may vary. Sample answer: Assume that air resistance is negligible. Dropping a ball
from the window of a moving car rather than stationary car should make no difference to the
time it takes for the ball to reach the ground. The motion of the car affects only the horizontal
motion of the ball. The time to fall is affected only by gravity.
33. Answers may vary. Sample answer: An object can be at rest as well as in motion at the same
time. It depends on who is observing the object. A bag of skis tied to the roof of a car is at rest as
seen by someone in the car. If the car is driving along the street, someone standing on the
sidewalk would say the skis are in motion. A full description of any object’s motion should
include mention of a frame of reference, the point of view from which the observations are made.
34. Answers may vary. Sample answer: To navigate a boat so it goes directly across a fastmoving river, I would need to head the boat somewhat upstream. If I have the correct angle, I
would observe that the boat is moving along a line directly across the river even though the bow
of the boat is not pointing that way. If I am slipping downstream, I need to point more upstream.
If I am moving upstream a bit, I need to point more to the opposite shore.
Analysis and Application
35. During interval P, the graph is horizontal, showing that the object is at rest. During interval
Q, the graph has a constant positive slope, showing that the object is moving forward at constant
speed. During interval R, the graph is horizontal, showing that the object is at rest. During
interval S, the graph has a constant negative slope, showing that the object is moving backward
at constant speed.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-4
36.
37. Answers may vary. Sample answer: A juggler is able to time the throw and the catch of the
balls to ensure they follow certain trajectories. If the juggler throws the balls at the same
velocities each time, then they will follow predictable paths. This allows the juggler to determine
where the balls will be when they need to be caught.
!
!
38. Given: vi = 0 m/s; !t = 2.0 s; a = 9.8 m/s 2 [down]
!
Required: !d
! !
1!
Analysis: The initial velocity, time taken, and acceleration are known. Use !d = vi !t + a!t 2 .
2
Use up as the positive direction.
1
Solution: !d = vi !t + a!t 2
2
1
!d = (0 m/s)(2.0 s) + ("9.8 m/ s 2 )(2.0 s)2
2
= "19.6 m
!d = "2.0 # 101 m
Statement: The squirrel was 2.0 ×101 m above the ground.
!
!
39. Given: vi = 9 m/s [forward]; vf = 0 m/s; !t = 3 s
!
Required: aav
!
! !v
Analysis: The initial and final velocities and the time taken are known. Use a =
. Use
!t
forward as the positive direction.
!v
Solution: a =
!t
0 m/s " 9 m/s
=
3s
a = "3 m/s 2
Statement: The runner’s acceleration is 3 m/s2 [backward].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-5
!
!
40. (a) Given: vi = 0 m/s; vf = 9.0 m/s [forward]; !t1 = 2.0 s
!
Required: a
!
! !v
Analysis: a =
; use forward as positive.
!t
!v
Solution: a =
!t
9.0 m/s " 0 m/s
=
2.0 s
a = 4.5 m/s 2
Statement: The runner’s acceleration is 4.5 m/s2 [forward].
r
r
(b) Given: total displacement: vi = 0 m/s; vf = 9.0 m/s [forward]; !t1 = 2.0 s;
r
!d = 100.0 m [forward]; second part of race: v = 9.0 m/s [forward]; !t1 = 2.0 s
Required: total time, !t
Analysis: The total displacement of the runner is 100.0 m but the actual running of the race has
! !
!
1!
two parts. Use !d = vi !t + a!t 2 to calculate the displacement !d1 during the acceleration of
2
!
the first part of the race. Then calculate the displacement !d2 for the second part of the race and
!
!d
!
use the constant velocity formula, vav =
, to determine the time !t2 for the second part of the
!t
!d
race; !t =
. Finally, the total time taken is the sum of the two times, !t = !t1 + !t2 .
vav
Solution: The displacement for the first part of the race is
1
!d1 = vi !t + a!t 2
2
1
= (0 m/s )(2.0 s) + (4.5 m/ s 2 )(2.0 s)2
2
!d1 = 9.0 m
The displacement for the second part of the race is
!d = !d1 + !d2
!d2 = !d " !d1
= 100.0 m " 9.0 m
!d2 = 91.0 m
The time for the second part of the race is
!d
!t2 = 2
v2
91.0 m
9.0 m/s
!t2 = 10.1 s
=
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-6
The time for the entire race is
!t = !t1 + !t2
= 2.0 s + 10.1 s
!t = 12 s
Statement: The runner takes 12 s to run the race.
!
!
!
41. (a) Given: v1 = 20.0 m/s [forward]; vf = 0 m/s; !d = 50.0 m
!
Required: a
Analysis: Use the initial and final velocities and the displacement. Use vf2 = vi2 + 2a!d to
determine the acceleration: a =
vf2 ! vi2
. Use forward as positive.
2"d
vf2 ! vi2
2"d
(0 m/s)2 ! (20.0 m/s)2
=
2(50.0 m)
Solution: a =
a = !4.00 m/s 2
Statement: The car’s acceleration is 4.00 m/s2 [backward].
!
!
!
(b) Given: v1 = 20.0 m/s [forward]; vf = 0 m/s; a = 4.00 m/s 2 [backward]
Required: stopping time, !t
r
r !v
!v
; !t =
Analysis: Any of the formulas involving !t can be used. Use a =
.
!t
a
!v
Solution: !t =
a
0 m/s " 20.0 m/s
=
"4.00 m/s 2
!t = 5.00 s
Statement: It takes the race car 5.00 s to stop.
!
42. Given: vi = 5.0 m/s [forward]; !d = 10.0 m [forward]; !t = 2.2 s
!
Required: a
r r
!d " vi !t
1r
Analysis: Use !d = vi !t + a!t 2 to calculate the acceleration: a =
.
2
0.5!t 2
Use forward as the positive direction.
!d " vi !t
Solution: a =
0.5!t 2
10.0 m " (5 m/ s )(2.2 s ) m
a=
0.5(2.2 s)2
a = "0.41 m/s 2
Statement: The bowling ball’s acceleration is 0.41 m/s2 [backward].
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-7
r
r
43. Given: stone 1: vi = 0 m/s; ti = 0 s; stone 2: vi = 10.0 m/s [down]; ti = 0.50 s
!
Required: displacement where stone 2 overtakes stone 1, !d
Analysis: Both stones accelerate down under the influence of gravity. Rewrite the equation
! !
1!
!d = vi !t + a!t 2 for each stone. Then compare the equations to determine the time when the
2
stones have the same displacement. Use up as the positive direction.
! !
1!
Solution: Rewrite the equation !d = vi !t + a!t 2 for each stone at time t (omitting the units for
2
convenience).
Stone 1:
1
!d = vi !t + a!t 2
2
1
= (0 m/s)(t " 0 s)+ ("9.8 m/ s 2 )(t " 0 s)2
2
2
!d = "4.9t
(Equation 1)
Stone 2:
1
!d = vi !t + a!t 2
2
1
= ("10 m/s)(t " 0.5 s) + ("9.8 m/ s 2 )(t " 0.5 s)2
2
= ("10t + 5.0) " (4.9)(t 2 " t + 0.25)
(Equation 2)
!d = "4.9t 2 " 5.1t + 3.775
Compare Equations 1 and 2 and solve for t.
!4.9t 2 = !4.9t 2 ! 5.1t + 3.775
5.1t = 3.775
t = 0.7402 s (two extra digits carried)
Substitute this value of t into either Equation 1 or 2 to determine the common displacement.
!d = "(4.9 m/s 2 )t 2
(Equation 1)
2
= "(4.9 m/ s )(0.7402 s )2
!d = "2.7 m
Statement: The second stone overtakes the first stone 2.7 m below the top of the cliff.
44. (a) Answers may vary. Sample answer: To see how the height of a high jumper changes
when the acceleration of gravity is reduced, I thought about a projectile problem where you are
determining the maximum height attained. For vertical initial velocity, the maximum height is
vi2
!d y =
, so a smaller value of g would lead to a higher jump. Another way to see this is to
2g
realize that it is the effect of gravity that stops you from jumping very high. The weaker the force
of gravity is, the higher a jump will be.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-8
!
(b) Given: vi = 5 m/s [up]; g = 2.0 m/s 2 [down]
Required: !d y
Analysis: !d y =
vi2
2g
Solution:
(5 m/s)2
!d y =
2(2.0 m/ s 2 )
!d y = 6.2 m
Statement: You could throw a baseball up 6.2 m.
!
45. Given: v = 60 m/s [S 60° W]
!
Required: vx
Analysis: Use east and north as positive. The component directed west is in the negative x
direction.
Solution: vx = !v sin "
= !(60 m/s)(sin60°)
vx = !50 m/s
Statement: The required component of velocity is 50 m/s [W].
46. Solutions may vary. Sample answer:
!
!
!
(a) Given: car: d1i = 0 m [forward]; v1i = 0 m/s; a = 2 m/s 2 [forward] ;
r
r
truck: d2i = 0 m [forward]; v2 = 10 m/s [forward]
Required: displacement where the car catches up with the truck, !d
Analysis: The vehicles start from the same position and travel an equal distance. Set up an
equation for each vehicle that relates position and time. The car moves at constant acceleration,
r
! !
r !d
1! 2
!d = vi !t + a!t , and the truck at constant velocity, v =
so !d = v!t . Compare these
2
!t
equations to solve for time and, later, displacement. Use forward as positive.
Solution:
truck:
car:
1
!d = vi !t + a!t 2
2
1
!d1 = (0 m/s)!t + (2 m/s 2 )!t 2
2
2
2
!d1 = (1 m/s )!t
The displacements of the two vehicles are equal,
!d1 = !d2
2
(1 m/s )!t 2 = (10 m/s) !t
!t = 10 s
Copyright © 2012 Nelson Education Ltd.
!d2
!t
!d2 = v2 !t
v2 =
!d2 = (10 m/s)!t
Chapter 1: Kinematics
1-9
The common displacement can be found using the equation for either the car or the truck.
!d1 = (1 m/s 2 )!t 2
= (1 m/ s 2 )(10 s)2
!d1 = 100 m
Statement: The car overtakes the truck 100 m beyond the starting point.
!
!
!
(b) Given: car: d1i = 0 m [forward]; v1i = 0 m/s; a = 2 m/s 2 [forward]
Required: speed of the car at 10 s, vf
r r
!
r vf ! vi r r r
! !v
; vf = vi + a"t
Analysis: Use a =
to solve for the final speed of the car; a =
!t
"t
Solution: vf = vi + a!t
= 0 m/s + (2 m/s 2 )(10 s)
vf = 20 m/s
Statement: The car is moving at 20 m/s [forward] when it catches up with the truck.
!
!
47. Given: vi = 100.0 km/h [forward]; vf = 0 m/s; !t = 5.2 s
!
Required: !d
! " v! + v! %
Analysis: Use !d = $ i f ' !t to solve for displacement. Use forward as the positive direction.
# 2 &
First convert kilometres per hour to metres per second.
1000 m
1 h
100.0 km
!
!
= 27.78 m/s (two extra digits carried)
3600 s
1 h
1 km
"v +v %
Solution: !d = $ i f ' !t
# 2 &
(27.78 m/s + 0 m/s)(5.2 s)
2
!d = 72 m
Statement: The car stops in 72 m.
!
!
!
48. (a) Given: vi = 30.0 m/s [up]; vf = 0 m/s; a = 9.8 m/s 2 [down]
Required: !t
!
!v
! !v
Analysis: Use a =
to calculate the time taken; !t =
. Use up as positive.
!t
a
!v
Solution: !t =
a
(0 m/s) " (30.0 m/s)
=
"9.8 m/s 2
!t = 3.1 s
Statement: It takes the ball 3.1 s to rise to its maximum height.
=
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-10
!
!
!
(b) Given: vi = 30.0 m/s [up]; vf = 0 m/s; a = 9.8 m/s 2 [down]
!
Required: !d
v 2 " vi2
Analysis: Use !d = f
. (Any of the formulas containing displacement could be used.)
2a
vf2 " vi2
Solution: !d =
2a
(0 m/s) 2 " (30.0 m/s) 2
=
2("9.8 m/ s 2 )
!d = 46 m
Statement: The ball rises 46 m.
!
!
!
(c) Given: vi = 30.0 m/s [up]; vf = 10.0 m/s [up]; a = 9.8 m/s 2 [down]
Required: !t
!
!v
! !v
Analysis: Use a =
to calculate the time taken; !t =
. Use up as positive.
!t
a
!v
Solution: !t =
a
(10.0 m/s) " (30.0 m/s)
=
"9.8 m/s 2
!t = 2.0 s
Statement: At 2.0 s after the throw, the ball will have a velocity of 10.0 m/s [upward].
!
!
!
(d) Given: vi = 30.0 m/s [up]; vf = 10.0 m/s [down]; a = 9.8 m/s 2 [down]
Required: !t
!
!v
! !v
Analysis: Use a =
to calculate the time taken; !t =
. Use up as positive.
!t
a
!v
Solution: !t =
a
("10.0 m/s) " (30.0 m/s)
=
"9.8 m/s 2
!t = 4.1 s
Statement: At 4.1 s after the throw,
will have a velocity of 10.0 m/s [downward].
r the ball
r
r
(e) Given: vi = 30.0 m/s [up]; !d = 0 m; a = 9.8 m/s 2 [down]
Required: !t
Analysis: The displacement is zero when the ball returns to its original height. Its velocity at this
!
! !v
!
time is the opposite of its initial velocity: vf = 30.0 m/s [down] . Use a =
to calculate the
!t
time taken. Use up as positive.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-11
!v
!t
!v
!t =
a
("30.0 m/s) " (30.0 m/s)
=
"9.8 m/s 2
!t = 6.1 s
Statement: The displacement is zero after 6.1 s.
49. (a) Given: velocity–time graph Figure 2, page 56
Required: instantaneous acceleration at t = 3 s, t = 10 s, and t = 13 s
Analysis: The velocity–time graph is straight at the requested times. The instantaneous
acceleration at these times is equal to the average acceleration close to these times. For each
!
!v
!
time, read two points from the graph and use aav =
. Use south as the positive direction.
!t
Solution: For t = 3 s, two points are (0 s, 20 m/s) and (5 s, 20 m/s).
!v
aav =
!t
20 m/s " 20 m/s
=
5 s"0 s
aav = 0 m/s 2
Solution: a =
For t = 10 s, two points are (9 s, 45 m/s) and (12 s, 15 m/s).
!v
aav =
!t
15 m/s " 45 m/s
=
12 s " 9 s
aav = "10 m/s 2
For t = 13 s , two points are (12 s, 15 m/s) and (14 s, 5 m/s).
!v
aav =
!t
5 m/s " 15 m/s
=
14 s " 12 s
aav = "5 m/s 2
Statement: The instantaneous acceleration at t = 3 s is 0 m/s2; at t = 10 s, it is 10 m/s2 [N]; and at
t = 10 s, it is 5 m/s2 [N].
(b) Given: velocity–time graph Figure 2, page 56
!
Required: average acceleration for the complete motion, aav
Analysis: The average acceleration is based only on the change in velocity from the beginning to
!
!v
!
the end of the motion. Read the initial the final points from the graph and use aav =
. Use
!t
south as the positive direction.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-12
Solution: The two points are (0 s, 20 m/s) and (16 s, 0 m/s).
!v
aav =
!t
0 m/s " 20 m/s
=
16 s " 0 s
aav = "1.2 m/s 2
Statement: The average acceleration for the complete motion is 1.2 m/s2 [N].
50. Given: At t1 = 2 s and t2 = 10 s, d y = h; g = 9.8 m/s 2
Required: height, h
Analysis: The ball’s flight is symmetric on the way up and the way down. The time the ball
reaches its maximum is halfway between the two given times. Use this time and !v y = "g!t to
1
determine the initial vertical velocity. Then use !d y = v1y !t " g!t 2 with t = 2 s, to determine
2
the height h. Use up as the positive direction.
Solution: The time of maximum height is
t +t
!t = 1 2
2
2 s + 10 s
=
2
!t = 6 s
Calculate the initial vertical velocity v1y .
!v y = "g!t
v yf " v yi = "g!t
0 m/s " v yi = ("9.8 m/s 2 )(6 s)
v yi = 58.8 m/s
The displacement after 2 s is
1
!d y = v1y !t " g!t 2
2
= (58.8 m/s)(2 s) " (4.9 m/ s 2 )(2 s) 2
= 117.6 m " 19.6 m
!d y = 98 m
Statement: The height at 2 s and at 10 s is 98 m.
! !
1!
51. !d = vi !t + aav !t 2
2
m
m 2
[m] = [ s ] +
[s ]
s
s2
[m] = [m] + [m]
[m] = [m]
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-13
! !
1!
Since both sides of !d = vi !t + aav !t 2 have the SI units of distance, the equation is
2
dimensionally correct.
! !
! " v! + v! %
1!
2
2
52. (a) To derive vf = vi + 2a!d from !d = $ i f ' !t and !d = vi !t + a!t 2 , isolate !t in
2
# 2 &
the first formula and substitute this into the second formula, then solve for vf2 . Since these
!
formulas apply to motion in one dimension, use the convention that the vector w is renamed w,
with the sign of w indicating direction.
"v +v %
Isolate !t in !d = $ i f ' !t .
# 2 &
"v +v %
!d = $ i f ' !t
# 2 &
!t =
2!d
v i + vf
1
Substitute this into !d = vi !t + a!t 2 .
2
1
!d = vi !t + a!t 2
2
" 2!d % 1 " 2!d %
!d = vi $
' + a$
'
# vi + vf & 2 # v i + vf &
2
" vi %
" 1 %
+ 2a!d 2 $
!d = 2 !d $
'
'
# vi + vf &
# vi + vf &
" vi %
" 1 %
+
2a!d
1 = 2$
'
$v +v '
# vi + vf &
# i f&
2
2
Multiply both sides by (vi + vf )2 and simplify.
(vi + vf )2 = 2vi (vi + vf ) + 2a!d
vi2 + 2vivf + vf2 = 2vi2 + 2vivf + 2a!d
vf2 = vi2 + 2a!d
! !
! !
! " v! + v! %
1! 2
1!
!
(b) To derive !d = vf !t " a!t from !d = $ i f ' !t and !d = vi !t + a!t 2 , eliminate vi .
2
2
# 2 &
!
Isolate vi in the first formula, substitute into the second formula, and then rearrange the result to
obtain the required formula.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-14
Isolate vi .
"v +v %
!d = $ i f ' !t
# 2 &
v i + vf =
2!d
!t
2!d
!t
Substitute and rearrange.
1
!d = vi !t + a!t 2
2
#
2!d &
1
= % "vf +
!t + a!t 2
(
!t '
2
$
vi = (vf +
1
!d = "vf !t + 2!d + a!t 2
2
1
"!d = "vf !t + a!t 2
2
1
!d = vf !t " a!t 2
2
53. Answers may vary. Sample answer: I would use a long straight plank and a pile of textbooks
arranged on the floor or on a long counter. I would also need a ball and a ranger with its
computer or graphing calculator. I would place one end of the plank on some of the books, set
the ranger at the top of the plank, and release the ball from in front of the ranger to roll down the
plank. Using the software that comes with the ranger, I would draw position–time, velocity–time,
and acceleration–time graphs. I would examine the graphs to see if the ball underwent uniform
acceleration and, if so, determine the value of the acceleration. It would be interesting to change
the number of books under the plank and see how the acceleration changes. Here I would be
using the angle of inclination of the plank, as measured from its geometry, as the independent
variable and the acceleration of the ball as the dependent variable. For a fixed angle of
inclination, I might also compare the acceleration of a slippery sliding object with that of the
ball.
54. (a) The sign of the slope of the velocity–time graph gives the sign of the acceleration. The
acceleration is positive between 0 s and 10 s, and again between 40 s and 50 s. The acceleration
is negative between 20 s and 40 s. The acceleration is 0 m/s2 between 10 s and 20 s.
(b) The initial velocity and final velocity are the same, so the average acceleration is 0 m/s2.
(c) The object changes direction when its velocity reverses or changes sign. The object changes
direction at 30 s.
(d) Given: velocity–time graph, Figure 3, page 57
Required: displacement between 0 s and 10 s, !d1 ; displacement between 10 s and 15 s, !d2
Analysis: Displacement is the area under (or above) the velocity–time curve between the given
times. The first displacement is calculated by looking at a triangle; the second is found using a
rectangle. Use west as the positive direction.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-15
Solution:
From 0 s and 10 s,
1
!d1 = (base)(height)
2
1
= (10 s)(60 m/s )
2
!d1 = 300 m
From 10 s and 15 s,
!d2 = (base)(height)
= (5 s)(60 m/s )
!d2 = 300 m
Statement: Both displacements are 300 m [W].
55. (a) Given: vx = 10.0 m/s; viy = 0 m/s; !d y = "0.49 m
Required: !d x
Analysis: In the horizontal direction, I know the constant speed but I do not know the time
interval or the displacement. In the vertical direction, I know the displacement, initial velocity,
1
and acceleration. Determine the time interval using !d y = v1y !t " g!t 2 ;
2
1
v1y ± v1y2 " 4( g)(!d y )
2
!t =
g
Then go back to the horizontal direction and use !d x = vx !t to determine the required horizontal
distance.
Solution:
vertical direction:
1
v1y ± v1y2 " 4( g)(!d y )
2
!t =
g
0 m/s ± (0 m/s)2 " 4(4.9 m/s 2 )("0.49 m)
9.8 m/s 2
!t = ±0.3162 s
!t =
horizontal direction:
!d x = vx !t
= (10.0 m/s )(0.3162 s)
!d x = 3.2 m
Statement: The dart board and player are 3.2 m apart.
r
r
r
56. Given: B = 5.0 units [along +x-axis]; A = 5.0 units [at 45° to B]
! !
Required: components of A ! B
Analysis: Make a sketch of the two vectors.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-16
!
!
!
!
A
B
that
points
above
, into the first
It is not clear whether A points above or below B!. Assume
!
!
A
B
B
quadrant. Determine
the
xand
y-components
of
and
.
Then
subtract
the
components
of
!
from those of A .
Solution:
!
!
The components of A are
The components of B are
!
!
Bx = B sin !
Ax = A sin !
Bx = 5.0 units
= (5.0 units)(sin 45°)
!
Ax = 3.535 units (two extra digits carried)
B
=
B
cos!
y
!
Ay = A cos!
By = 0 units
= (5.0 units)(cos 45°)
Ay = 3.535 units (two extra digits carried)
! !
The components of A ! B are
! !
!
!
! !
!
!
( A ! B) y = Ay ! By
( A ! B) x = Ax ! Bx
! ! = 3.535 units ! 0 units
! ! = 3.535 units ! 5.0 units
( A ! B) y = +3.5 units
( A ! B) x = !1.5 units
! !
A
! B points left and up.
Looking at the sketch above, this makes sense
because
! !
Statement: The x- and y-components of A ! B are !1.5 units and +3.5 units respectively.
!
!
57. Given: !d1 = 20.0 km [N]; !d2 = 25.0 km [N 60.0° W]
!
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 . Determine the x- and y-components of the given displacement
vectors. Add these x- and y-components to determine the x- and y-components of the total
displacement. Finally, use the Pythagorean theorem and tangent ratio to determine the total
displacement vector. Use east and north as positive.
Solution: For the first vector,
!
!d1x = !d1 sin "
= +(20.0 km)(cos 90.0°)
!d1x = 0 km
!
!d1y = + !d1 cos"
= +(20.0 km)(sin 90.0°)
!d1y = +20.0 km
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-17
For the second vector,
!
!d2x = " !d2 sin #
= "(25.0 km)(sin 60.0°)
!d2x = "21.651 km (two extra digits carried)
!
!d2y = + !d2 cos"
= +(25.0 km)(cos 60.0°)
!d2y = +12.5 km
Add the horizontal components.
!d x = !d1x + !d2x
= 0 km + ("21.651 km)
!d x = "21.651 km (two extra digits carried)
Add the vertical components.
!d y = !d1y + !d2y
= (+20.0 km) + (+12.5 km)
!d y = +32.5 km
Combine the total displacement components to determine the total displacement.
!
!d = !d x2 + !d y2
=
( 21.651 km )2 + ( 32.5 km )2
!
!d = 39.1 km
!
$ #d
y
! = tan "1 & !
&% #d x
'
)
)(
$ 32.5 km '
= tan "1 &
)
% 21.651 km (
! = 56.3°
Statement: The total displacement of the trip is 39.1 km [W 56.3° N] or 39.1 km [N 33.7° W].
!
!
58. (a) Given: !d1 = 6.0 m [positive y-axis]; !d2 = 8.0 m [23° below positive x-axis]
!
Required: !d
!
!
!
Analysis: !d = !d1 + !d2 Determine the x- and y-components of the given displacement vectors.
Add these x- and y-components to determine the x- and y-components of the total displacement.
Finally, use the Pythagorean theorem and the tangent ratio to determine the total displacement
vector.
Solution: For the first vector,
!
!d1x = !d1 cos"
= (6.0 m)(cos 90°)
!d1x = 0 m
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Chapter 1: Kinematics
1-18
!
!d1y = + !d1 sin "
= +(6.0 m)(sin 90°)
!d1y = +6.0 m
For the second vector,
!
!d2x = + !d2 cos"
= +(8.0 m)(cos 23°)
!d2x = 7.364 m (two extra digits carried)
!
!d2y = " !d2 sin #
= "(8.0 m)(sin 23°)
!d2y = "3.126 m (two extra digits carried)
Add the x-components.
!d x = !d1x + !d2x
= 0 m + (+7.364 m)
= +7.364 m
!d x = +7.4 m
Add the y-components.
!d y = !d1y + !d2y
= (+6.0 m) + ("3.126 m)
= +2.874 m
!d y = +2.9 m
Statement: The x- and y-components of the total displacement vector are +7.4 m and +2.9 m.
(b) Given: !d x = +7.364 m; !d y = +2.874 m
!
Required: !d
Analysis: Combine the total displacement components to determine the total displacement.
!
Solution: !d = !d x2 + !d y2
!
2
2
!d = ( 7.364 m ) + ( 2.874 m )
!
!d = 7.9 m
!
$ #d '
y
! = tan "1 & ! )
&% #d x )(
$ 2.874 m '
= tan "1 &
% 7.364 m )(
! = 21°
Statement: The total displacement is 7.9 m at 21° above the positive x-axis.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-19
!
!
59. Given: vi = 15 m/s [E]; vf = 12 m/s [E 25° N]; !t = 5.0 s
!
Required: aav
! ! !
Analysis: Calculate the change in velocity using components and !v = vf " vi . Then, determine
!
!v
!
the average acceleration using aav =
. Use east and north as positive.
!t
!
Solution: The x- component of !v is
!vx = vfx " vix
= (12 m/s)(cos 25°) " 15 m/s
= 10.875 m/s " 15 m/s
!vx = "4.124 m/s
!
The y-component of !v is
!v y = vfy " viy
= (12 m/s)(sin 25°) " 0 m/s
!v y = +5.071 m/s
Determine the change in velocity from its components.
!
!v = !vx2 + !v 2y
=
( 4.124 m/s )2 + (5.071 m/s )2
!
!v = 6.5 m/s
$ #v!
y
"1
! = tan & !
&% #vx
'
)
)(
$ 5.071 m/s '
= tan "1 &
)
% 4.124 m/s (
= 50.88°
! = 51°
Calculate the average acceleration.
!
!v
!
aav =
!t
6.536 m/s [W 51° N]
=
5.0 s
!
2
aav = 1.3 m/s [W 51° N]
Statement: The average acceleration of the car is 1.3 m/s 2 [W 51° N] .
60. Answers may vary. Sample answer: Assume that air resistance is negligible. The golf ball is
in projectile motion so its horizontal motion is at constant speed. The average acceleration in the
x-direction is 0 m/s2.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-20
61. (a)
(b)
!
!
62. Given: !t = 6.0 s; !d = 60.0 m [forward]; vf = 15 m/s [forward]
Required: a; vi
2(vf !t " !d)
1
Analysis: Use !d = vf !t " a!t 2 to determine the acceleration: a =
.
2
!t 2
Any of the other motion formulas can be used to calculate the initial velocity. Use vf = vi + a!t .
Use forward as the positive direction.
Solution:
Solve for acceleration.
Solve for initial velocity.
2(vf !t " !d)
vf = vi + a!t
a=
!t 2
vi = vf " a!t
2((15 m/s)(6.0 s) " 60.0 m)
= 15 m/s " (1.667 m/s 2 )(6.0 s)
a=
2
(6.0 s)
= 15 m/s " 10 m/s
= 1.667 m/s 2 (two extra digits carried)
vi = 5 m/s
2
a = 1.7 m/s
Statement: The acceleration is 1.7 m/s2 [forward] and the initial speed is 5 m/s.
63. Solutions may vary. Sample solution:
Given: maximum height: !d y = 3.7 m; " = 45°; g = 9.8 m/s 2
Required: vi
Analysis: Imagine the puma completing its jump and returning to the ground. The puma’s jump
(v sin " )2
is a projectile motion problem. The maximum height of a projectile is !d y = i
.
2g
Solve for vi ; vi2 =
2g!d y
sin 2 "
.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-21
Solution: vi2 =
2g!d y
sin 2 "
2(9.8 m/s 2 )(3.7 m)
=
sin 2 45°
vi2 = 145.04 m 2 /s 2
vi = 12 m/s
Statement: The initial speed of the puma is 12 m/s.
64. Answers may vary. Sample answer:
(a) Imagine the footballs returning to the ground. There are three formulas developed for such
(vi sin " )2
symmetric projectile motion. The one for maximum height is !d y =
. If the two balls
2g
are kicked with the same initial speed, the one with the greater angle (A) goes higher.
2v sin "
. The football with the greater angle (A) stays
(b) The formula for time of flight is !t = i
g
in the air longer.
v 2 sin 2"
(c) No because neither angle is known. The formula for the horizontal range is !d x = i
.
g
Footballs kicked at complementary angles have the same range. Otherwise the football with the
angle closer to 45° would have the longer range.
65. Given: !d x = 200.0 m; " = 45°; diy = dfy ; g = 9.8 m/s 2
Required: vi
Analysis: Use the range formula, !d x =
vi2 sin 2"
.
g
vi2 sin 2"
g
!d x g
vi2 =
sin 2"
(200.0 m)(9.8 m/s 2 )
=
sin 90°
2
vi = 1960 m 2 /s 2
Solution: !d x =
vi = 44 m/s
Statement: The baseball leaves the bat at 44 m/s.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-22
66. Given: !d x = 9.5 m; d yi = 2.0 m; d yf = 3.1 m; " = 35°; g = 9.8 m/s 2
Required: vi
Analysis: I do not know the time interval for the shot or the initial speed of the ball. I do know
that vx = vi cos! and v y = vi sin ! . Write an equation for the horizontal motion using !d x = vx !t
1
and for the vertical motion using !d y = v1y !t " g!t 2 . Use these two equations to write a
2
quadratic equation for !t .
Solution: Horizontal motion:
!d x = vx !t
9.5 m = vi cos35°!t
9.5 m
(cos 35°)!t
Vertical motion:
vi =
(Equation 1)
1
!d y = v1y !t " g!t 2
2
(3.1 m " 2.0 m) = v1 (sin 35°)!t " (4.9 m/s 2 )!t 2
1.1 m = v1 (sin 35°)!t " (4.9 m/s 2 )!t 2 (Equation 2)
Substitute Equation 1 into Equation 2 and simplify.
" 9.5 m %
(sin 35° !t ) ( (4.9 m/s 2 )!t 2
1.1 m = $
# cos 35° !t '&
1.1 m = 6.652 m ( (4.9 m/s 2 )!t 2
6.652 m ( 1.1 m
4.9 m/s 2
!t = 1.064 s (two extra digits carried)
Finally, substitute this time interval into Equation 1.
9.5 m
vi =
cos 35°!t
9.5 m
=
(cos 35°)(1.064 s)
!t 2 =
vi = 11 m/s
Statement: The initial speed of the basketball is 11 m/s.
r
67. Given: vi = 30.0 m/s [45° above horizontal]; diy = dfy ; g = 9.8 m/s 2
Required: !t
Analysis: Since the ball lands at the same height from which it was launched, the time taken is
2v sin "
given by !t = i
.
g
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-23
2vi sin "
g
2(30.0 m/s )(sin 45°)
=
9.8 m/s 2
!t = 4.3 s
Statement: The time the ball stays in the air is 4.3 s.
r
68. (a) Given: vi = 60.0 m/s [60.0° above horizontal]; diy = dfy ; g = 9.8 m/s 2
Solution: !t =
Required: maximum height, d y
Analysis: The maximum height occurs when v y = 0 m/s . Use the acceleration formula
vf2y = viy2 ! 2g"d y to determine d y . Since viy = vi sin ! and vf y = 0 m/s , d y can be calculated.
vf2y = viy2 ! 2g"d y
"d y =
"d y =
vf2y
2g
(vi sin # )2
2g
Solution: !d y =
=
(vi sin " )2
2g
((60.0 m/s)sin 60.0°)2
2(9.8 m/s 2 )
!d y = 138 m
Statement: The maximum height of the stream of water is 138 m.
r
(b) Given: vi = 60.0 m/s [60.0° above horizontal]; g = 9.8 m/s 2
Required: horizontal range, !d x
vi2 sin 2"
Analysis: Use the range formula, !d x =
.
g
vi2 sin 2"
Solution: !d x =
g
(60.0 m/s)2 (sin 120.0°)
=
9.8 m/s 2
!d x = 318 m
Statement: The water hits the ground 318 m away.
69. Given: vx = 8.0 m/s; ! = 60.0°
Required: v y
Analysis: From basic trigonometry, v y = vx tan ! .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-24
Solution:
v y = vx tan !
= (8.0 m/s)( tan 60.0°)
v y = 14 m/s
Statement: The vertical component of the dolphin’s velocity is 14 m/s.
70. Answers may vary. Sample answer: The tennis player could have either hit the ball such that
it had a horizontal speed greater than 28.0 m/s or hit the ball at a different angle so the vertical
component of its velocity was greater.
71. Answers may vary. Sample answer: The long jumper can increase the length of his jump by
jumping at an angle closer to 45°.
r
72. (a) Given: vi = 26 m/s [75° above horizontal]; diy = dfy ; g = 9.8 m/s 2
Required: !d x
Analysis: Since the snowball lands at the height it was thrown from, use the range formula,
v 2 sin 2"
!d x = i
.
g
Solution: !d x =
vi2 sin 2"
g
(26 m/s)2 (sin 150°)
9.8 m/s 2
!d x = 34 m
Statement: The horizontal range of the snowball is 34 m.
(b) Since the two snowballs have the same initial speed and the same range, their angles of
inclination must be complementary. The second snowball is thrown at 15° to the horizontal.
r
r
(c) Given: vi1 = 26 m/s [75° above horizontal]; vi2 = 26 m/s [15° above horizontal]; g = 9.8 m/s 2
Required: time delay before throwing second snowball
2v sin "
Analysis: Calculate the time of flight for each snowball using !t = i
. The required time
g
delay between the throws is the difference in these times of flight.
Solution: snowball 1:
snowball 2:
2v sin "
2v sin "
!t = i1
!t = i2
g
g
2(26 m/s )(sin75°)
2(26 m/s )(sin15°)
=
=
2
9.8 m/s
9.8 m/s 2
!t1 = 5.125 s (two extra digits carried)
!t2 = 1.373 s (two extra digits carried)
The time delay should be
!t = !t1 " !t2
=
= 5.125 s " 1.373 s
!t = 3.8 s
Statement: He should throw the second snowball 3.8 s after throwing the first.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-25
r
73. (a) Given: vi = 27 m/s [20.0° above horizontal]; diy = dfy ; g = 9.8 m/s 2
Required: maximum height, d y
(vi sin " )2
Analysis: The maximum height occurs when v y = 0 m/s . Use the formula !d y =
to
2g
determine d y .
Solution: !d y =
=
(vi sin " )2
2g
((27 m/s)sin 20.0°)2
2(9.8 m/s 2 )
!d y = 4.4 m
Statement: The ball’s maximum height is 4.4 m above the ground.
(b) When air resistance is negligible, the flight is symmetric going up and coming down. The
speed of the ball as it hits the ground is 27 m/s.
74. (a) Given: vx = 27 m/s; viy = 0 m/s; !d y = "2.4 m; g = 9.8 m/s 2
Required: !t
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
1
Analysis: Use !d y = v1y !t " g!t 2 to determine !t : !t =
g
2
#1 &
v1y ± v1y2 " 4 % g ( (!d y )
$2 '
Solution: !t =
g
0 m/s ± (0 m/s)2 " 4(4.9 m/s 2 )("2.4 m)
9.8 m/s 2
!t = ±0.70 s
Statement: The volleyball hits the floor after 0.70 s.
(b) Given: vx = 27 m/s; !t = 0.6999 s
!t =
Required: !d x
Analysis: !d x = vx !t
Solution: !d x = vx !t
= (27 m/s )(0.6999 s)
!d x = 19 m
Statement: The volleyball travels a horizontal distance of 19 m.
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Chapter 1: Kinematics
1-26
(c) Given: viy = 0 m/s; !t = 0.6999 s; g = 9.8 m/s 2
Required: vf
Analysis: The horizontal component of the velocity is constant throughout. The vertical
component changes with constant acceleration: !v y = "g!t . Determine the vertical component
of the final velocity and then construct the final speed
Solution:
Determine the vertical component of the final velocity.
!v y = "g!t
vfy = viy " g!t
= 0 m/s " (9.8 m/s 2 )(0.6999 s)
vfy = "6.585 m/s (two extra digits carried)
Since vfx = vx = 1.93 m/s , combine the components to determine the final speed.
!
vf = (vfx )2 + (vfy )2
= (27 m/s)2 + (!6.585 m/s)2
!
vf = 28 m/s
Statement: The final speed of the volleyball is 28 m/s.
!
!
75. Given: vBW = 12.0 km/h [N]; vWG = 6.00 km/h [E]
!
Required: vBG
!
!
!
Analysis: The vector addition diagram for vBG = vBW + vWG is a right-angled triangle. Use the
!
Pythagorean theorem and the tangent ratio to determine vBG .
Solution:
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-27
!
Determine the magnitude of vBG .
vBG = (vBW )2 + (vWG )2
= (12.0 km/h)2 + (6.00 km/h)2
vBG = 13.4 km/h
!
Determine the direction of vBG
!
# vWG &
"1
! = tan % ! (
$ vBW '
# 6.00 km/h &
= tan "1 %
(
$ 12.0 km/h '
! = 26.6°
Statement: The velocity of boat with respect to the ground is 13.4 km/h [N 26.6° E].
76. (a) Given: swimmer leg 1: !t1 = 0.25 h; swims upstream ;
swimmer leg 2: swims downstream, meets raft; Raft: drifts downstream !d R = 1.0 km
Required: speed of current, w
Analysis: There are three different motions that interconnect: the speed of the current, w; the
speed of the swimmer with respect to the water, v; and the swimmer’s speed with respect to the
shore, v1 . For the swimmer going upstream, I know the time interval !t1 . The swimmer’s speed
with respect to the shore is v1 = v ! w . Call the distance covered !d1 .
For the swimmer going downstream, his speed with respect to the water is v2 = v + w . Call the
time taken !t2 and the distance covered !d2 .
For the raft, the distance covered is !d R = 1.0 km and its speed is vR . Call the time taken
!tR = t . All of the motions occur at constant speed, so rewrite the relation !d = v!t for each
motion. Units will be omitted for clarity and convenience. At this point some algebraic
manipulation should yield the value of w.
Solution:
swimmer leg 1:
swimmer leg 2:
raft:
!d2 = v2 !t2
!d R = vR !tR
!d1 = v1!t1
!d1 =
(v " w)
4
# 1&
!d2 = (v + w) % t " (
$ 4'
The swimmer swims to meet the raft, so
!d2 = !d1 + !d R
Substitute into this last equation and simplify.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1 = wt
t=
1
w
1-28
!d2 = !d1 + !d R
# 1 & (v " w)
(v + w) % t " ( =
+1
$ 4'
4
# 1 1 & (v " w)
(v + w) % " ( =
+1
$ w 4'
4
# 4 " w & (v " w)w 4w
(v + w) %
+
=
$ 4w ('
4w
4w
(v + w)(4 " w) = (v " w)w + 4w
4v + 4w " vw " w2 = vw " w2 + 4w
4 v = 2 vw
w=2
Statement: The speed of the current is 2.0 km/h.
!
!
77. Given: vSW = 0.45 m/s [N]; vWG = 2.5 m/s [W]; !t = 200.0 s
Required: width of river, !d
!
Analysis: Since the swimmer heads directly across the river, her velocity, vSW , is actually the
!
[N]-component of vSG . Just use !d = v!t and calculate !d . Note that the size of the current is
irrelevant in this problem.
Solution: !d = v!t
= (0.45 m/s)(200.0 s)
!d = 9.0 " 101 m
Statement: The river is 9.0 × 101 m wide.
!
78. Given: !d = 35 m; !t = 4.0 min = 240 s; vWG = 0.25 m/s [parallel to shore];
!
vBG is [perpendicular to shore]
!
Required: vBW
Analysis: Use the given time interval and displacement to determine the speed of the boat with
!
!
!
respect to the ground. The vector addition diagram for vBG = vBW + vWG is a right-angled triangle.
!
Use the Pythagorean theorem and the tangent ratio to determine vBW .
!
Solution: Determine the magnitude of vBG .
!d
!t
35 m
=
240 s
= 0.1458 m/s
vBG =
vBG
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-29
!
Determine the magnitude of vBW .
vBW = (vWG )2 + (vBG )2
= (0.25 m/s)2 + (0.1458 m/s)2
vBW = 0.29 m/s
!
Determine the direction of vBW
!
# vWG &
"1
! = tan % ! (
$ vBE '
# 0.25 m/s &
= tan "1 %
(
$ 0.1458 m/s '
! = 60°
Statement: The velocity of the boat with respect to the ground is 0.29 m/s at 60° upstream from
straight across.
!
!
79. Given: vPA = 290 km/h [E 42° S]; vAG = 65 km/h [E 25° N]
!
Required: vPG
Analysis: Use the component method of solution. Use (vPG ) x = (vPA ) x + (vAG ) x and
(vPG ) y = (vPA ) y + (vAG ) y , with east and north as positive.
Solution: x-components:
(vPG ) x = (vPA ) x + (vAG ) x
= (290 km/h)(cos 42°) + (65 km/h)(cos 25°)
= 215.51 km/h + 58.91 km/h
(vPG ) x = 274.42 km/h
y-components:
(vPG ) y = (vPA ) y + (vAG ) y
= (!290 km/h)(sin 42°) + (65 km/h)(sin 25°)
= !194.05 km/h + 27.47 km/h
(vPG ) y = !166.58 km/h
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-30
!
Now use these components to determine vPG :
r
vPG = (vPGx )2 + (vPGy )2
= (274.42 km/h)2 + (166.58 km/h)2
r
vPG = 320 km/h
# 166.58 km/h &
! = tan "1 %
(
$ 274.42 km/h '
! = 31°
Statement: The velocity of the plane relative to the ground is 320 km/h [E 31° S].
Evaluation
80. Answers may vary. Sample answer: The time for both balls to reach the ground is equal to
2h / g , where h is the height they are dropped from. This equation does not depend on the
mass of the balls, so the two balls will take the same amount of time to reach the ground.
81. Answers may vary. Sample answer: Navigation of space probes requires using displacement,
velocity, and acceleration vectors. Total displacement and relative-velocity calculations involve
the addition and subtraction of vectors.
82. Answers may vary. Sample answer: The argument is not correct because the condition of the
javelin being in the air longer will not always result in it travelling a greater distance. This is
because the time of flight depends on the vertical component of the initial velocity. If the javelin
stays in the air for a longer duration of time, it will not necessarily cover a greater horizontal
range. To be able to throw the javelin a greater distance, the thrower should also consider the
speed and angle at which the javelin is thrown.
83. Answers may vary. Sample answer: If there were no gravity, the arrow would follow a
straight-line trajectory and the archer should aim directly at the bull’s-eye. Including the effect of
gravity means the arrow follows a parabolic trajectory, not a straight line. To hit the bull’s-eye,
the archer should give the arrow’s speed a small vertical component, ensuring that by the time it
reaches the target, it will be at the right height to hit the bull’s-eye. This small vertical
component is achieved by aiming higher than the bull’s-eye.
84. Answers may vary. Sample answer: The golfer can increase the range by increasing the angle
of flight of the ball. This ensures that the ball stays in the air for longer. Since the wind is
blowing from behind, this will carry the ball a greater distance.
85. Answers may vary. Sample answer: The equations will remain the same, but the values for
time of flight, horizontal range, and maximum height will be smaller for planet A; for planet B,
the equations will remain the same and the values will all be larger.
86. Answers may vary. Sample answer: The range of the snowball will be larger than 81 m. The
friend’s answer comes from using the range formula for an object landing at the same height as it
was launched. In this case, the snowball was launched from a height of 45 m, returned to 45 m
and landed on the ground. To determine the range of the snowball, you would have to solve this
more general problem.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-31
Reflect on Your Learning
87. Answers may vary. Sample answer: Even though I studied falling objects in grade 11, I keep
thinking about a falling object speeding up for a bit and then falling at constant speed. I have
realized this comes from my daily experience where many falling things experience lots of air
resistance. It takes a while to automatically picture the idealized situation as a starting point and
then add more realistic details as needed. I sometimes find it difficult to see the difference
between similar scalar and vector quantities such as speed and velocity or distance and
displacement. Drawing diagrams, especially in two dimensions, and actually following the
difference between distance and displacement was useful.
88. Answers may vary. Sample answer: Our work discussed a lot of projectile motion and
relative velocities. I realize that what we did is like a skeleton for solving more complex
problems. Certainly air resistance must come into projectile motion somehow. Launching a
rocket is a projectile problem, but the wind, the round and spinning Earth, decreasing air density,
and all sorts of complications need to be included. Airplanes navigating across the continent will
need to compensate for wind speeds and directions that change continuously and for how their
own airspeed depends on the load and type of fuel they are using. It feels good to begin to
separate the simple basic picture from the more realistic situations that need much more
experience.
89. Answers may vary. Sample answer: The strobe images like Figure 2 on page 37 and the ones
we made on the air table really showed me that falling objects don’t immediately move quickly.
Their speed really does build up in time. These images also convinced me that the horizontal
motion has no effect on the vertical acceleration.
90. Answers may vary. Sample answer: Looking at projectile motion helped me understand the
factors that affect the range of thrown ball. Knowing to throw a basketball at an angle close to
45° will help me make a better long pass.
Research
91. Answers may vary. Students’ answers should include some of the following information.
370: Aristotle claims that free falling bodies accelerate but heavier bodies fall faster.
1589: Galileo demonstrates that objects fall at the same rate independent of mass.
1604: Galileo discovers that the distance for falling objects increases as square of time.
1613: Galileo discovers the principle of inertia.
1637: Descartes formalizes the principle of inertia.
1684: Newton discovers the inverse square law and mass dependence of gravity.
1687: Newton publishes the laws of motion and gravitation.
92. Answers may vary. Students should describe how the ramps and turns on the track are
designed and what factors are taken into consideration. They can create a schematic diagram of
the dirt track model, conduct experiments and prepare data tables and validate various theories
they have learned in the chapter. Sample answer: On dirt bike tacks, the turns are banked so that
the bikers can go as fast as possible through the turns. Changing the angle of the bank will
change the velocity of the bikers as they pass through the turn. If turns are not banked, bikers
must decrease their acceleration in order to pass through them. For the jumps, the trajectory
depends on the angle of the jump and the velocity of the bikers. If the jump is angled at 45°, the
bikers will jump the farthest.
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-32
93. Answers may vary. The details of the report depend on the sport chosen. Students should
describe the motion of the objects that are involved in the sport. If possible, graphs may be
drawn to represent critical ideas. Sample answer: In the sport of football, accuracy is very
important in throwing and kicking the ball. Quarterbacks and punters try to throw and kick the
ball so that it spirals, which reduces the air drag on the ball allowing them to throw and kick the
ball farther and also more accurately. When quarterbacks and punters throw or kick the ball, the
ball follows a parabolic path, otherwise known as projectile motion. Since accuracy is very
important, the quarterback can change the trajectory of the ball by increasing or decreasing the
force that is applied to the ball or by changing the angle that the ball is thrown. To throw the ball
as far as possible, the quarterback should apply the maximum possible force and throw the ball at
a 45° angle, while throwing a perfect spiral. For punters, accuracy is not always important. Most
of the time punters want to kick the ball as far as possible and so they should apply the maximum
possible force and kick the ball at a 45° angle, while kicking the ball so it travels in a perfect
spiral. When accuracy does matter, such as when they want to determine their opponent’s field
position, like quarterbacks, the punter can change the trajectory of the ball by increasing or
decreasing the force that is applied to the ball or by changing the angle that the ball is kicked.
94. Answers may vary. Sample answer:
Distance (m)
Cumulative
Time for
Speed (m/s)
Time (s)
Interval (s)
10
1.85
1.85
5.41
20
2.87
1.02
9.80
30
3.78
0.91
11.0
40
4.65
0.87
11.5
50
5.50
0.85
11.8
60
6.32
0.82
12.2
70
7.14
0.82
12.2
80
7.96
0.82
12.2
90
8.79
0.83
12.0
100
9.69
0.90
11.1
This table shows the split times for Usain Bolt’s record-breaking run in the 2008 Olympics. Bolt
accelerated to a speed around 12 m/s, which he maintained for about 40 m, before decreasing his
acceleration at the end of the race. The acceleration was largest early in the race because this is
when the time for interval changed the most. From 50 m to 80 m, Bolt’s speed was exactly the
same at 12.2 m/s. Near the end of the race, Bolt started to slow down. Due to fatigue, any runner
would begin to slow down near the end; however, because Bolt had a sizable lead, he began to
celebrate before crossing the finish line, which slowed him down, so it is difficult to determine
how much of this decrease in speed was due to fatigue.
95. Answers may vary. Sample answer:
(a) Given: Reaction time v. Number of beers graph; possible initial speeds
Required: corresponding reaction distances
Analysis: A driver continues to drive at constant speed during the reaction time. Then braking
begins. The reaction distance is the distance travelled before the brakes are applied. For each
situation below, read the reaction time from the graph and then calculate the reaction distance
using !d = v!t .
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-33
Solution: The reaction time for “No alcohol” is 0.8 s. At 17 m/s, the reaction distance is
!d = v!t
= (17 m/ s )(0.8 s )
!d = 14 m
At 25 m/s, the reaction distance is
!d = v!t
= (15 m/ s )(0.8 s )
!d = 20 m
At 33 m/s, the reaction distance is
!d = v!t
= (33 m/ s )(0.8 s )
!d = 26 m
The reaction time for 4 bottles of beer is 2.0 s. At 17 m/s, the reaction distance is
!d = v!t
= (17 m/ s )(2.0 s )
!d = 34 m
At 25 m/s, the reaction distance is
!d = v!t
= (25 m/ s )(2.0 s )
!d = 50 m
At 33 m/s, the reaction distance is
!d = v!t
= (33 m/ s )(2.0 s )
!d = 66 m
The reaction time for 5 bottles of beer is 3.0 s. At 17 m/s, the reaction distance is
!d = v!t
= (17 m/ s )(3.0 s )
!d = 51 m
At 25 m/s, the reaction distance is
!d = v!t
= (25 m/ s )(3.0 s )
!d = 75 m
At 33 m/s, the reaction distance is
!d = v!t
= (33 m/ s )(0.8 s )
!d = 99 m
Copyright © 2012 Nelson Education Ltd.
Chapter 1: Kinematics
1-34
Table 2
Speed
17 m/s (60 km/h)
25 m/s (90 km/h)
33 m/s (120 km/h)
No alcohol
14
20
26
Reaction Distance (m)
4 bottles of beer 5 bottles of beer
34
51
50
75
66
99
(b) Answers may vary. Sample answer: Alcohol interferes with the transmission of the signals
from nerve cells to the brain. This results in slower reaction times because alcohol impairs
comprehension and coordination, in particular. The reaction time increases when the amount of
alcohol in the blood stream increases.
(c) Answers may vary. Sample answer: There is no safe level of alcohol that can be consumed by
drivers because alcohol affects people differently. People metabolize alcohol at different rates,
and many factors, such as one’s metabolic rate and body mass, affect the rate of metabolism. The
data in Table 2 show that at faster speeds, the distance travelled before brakes are applied
increases. Alcohol affects many other abilities, such as concentration, judgment, and vision,
which are very important for driving safely and other activities.
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Chapter 1: Kinematics
1-35