G
(d) vav = ?
G
∆d
G
vav =
∆t
65 m [43° S of E]
=
120 s
G
vav = 0.54 m/s [43° S of E]
The average velocity is 0.54 m/s [43° S of E].
Applying Inquiry Skills
9. (a) Students can refer to the Learning Tip titled “The Image of a Tangent Line” on page 13 of the text to understand how
to use a plane mirror to check the accuracy of their tangents.
(b) One way to draw tangents accurately is to use the plane mirror method, as described in the Learning Tip. Another way
that is useful for a displacement-time graph of uniform acceleration is to draw the tangent parallel to an imaginary line
joining two points that are equal times away from the tangent time (e.g., at t = 3.5 s, draw the tangent parallel to the
line joining the points at t = 2.5 s and t = 4.5 s).
Making Connections
10. Use the equation d = vav∆t to complete the table.
Reaction Distance
Speed
17 m/s (60 km/h)
25 m/s (90 km/h)
33 m/s (120 km/h)
no alcohol
14 m
20 m
26 m
4 bottles
34 m
50 m
66 m
5 bottles
51 m
75 m
99 m
1.2 ACCELERATION IN ONE AND TWO DIMENSIONS
PRACTICE
(Page 20)
Understanding Concepts
1. All five examples could be units of acceleration.
2. (a) It is possible to have an eastward velocity with a westward acceleration. For example, a truck moving eastward whle
slowing down has a westward acceleration.
(b) It is possible to have acceleration when the velocity is zero. For example, at the instant that a ball tossed vertically
upward comes to a stop, its acceleration is still downward.
3. (a) When the flock’s acceleration is positive, the flock is moving south with increasing velocity.
(b) When the flock’s acceleration is negative, the flock is moving south with decreasing velocity.
(c) When the flock’s acceleration is zero, the flock is moving south with constant velocity.
G
4. vi = 0
G
vf = 9.3 m/s [fwd]
∆t = 3.9 s
G
aav = ?
G G
vf − vi
G
aav =
∆t
9.3 m/s [fwd] − 0
=
3.9 s
G
aav = 2.4 m/s 2 [fwd]
The runner’s average acceleration is 2.4 m/s2 [fwd].
16
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
5.
G
vi = 0
G
vf = 26.7 m/s [fwd]
G
a = 9.52 m/s2
(a) ∆t = ?
∆t =
=
G
G
vf − vi
G
aav
26.7 m/s − 0
9.52 m/s 2
∆t = 2.80 s
The Espace takes 2.80 s to achieve a speed of 26.7 m/s.
(b) v = ?
m 1 km 3600 s
v = 26.7 ×
×
s 1000 m
1h
v = 96.1 km/h
The speed is 96.1 km/h.
G G
? v −v
(c) ∆t = f G i
aav
L L
−
T
T
T=
 L 
 2
T 
?
? L  T2 
 
T = 

 T  L 
?
6.
7.
T=T
Thus, the equation is dimensionally correct.
G
aav =14 (km/h)/s [fwd]
∆t = 4.7 s
G
vi = 42 km/h [fwd]
G
vf = ?
G G
v − vi
G
aav = f
∆t
G G G
vf = vi + aav ∆t
= 42 km/h [fwd ] + (14 (km/h)/s [fwd ]) ( 4.7 s )
G
vf =108 km/h [fwd ]
Thus, the final velocity is 108 km/h [fwd].
G
aav = 1.37 × 103 m/s2
∆t = 3.12 × 10–3 s
G
vf =0 m/s
G
vi = ?
G G
G
v − vi
aav = f
∆t
G G G
vi = vf − aav ∆t
(
)(
= 0 m/s − 1.37 ×103 m/s 2 [W ] 3.12 × 10−2 s
G
vi = 42.8 m/s [E ]
)
The velocity of the arrow as it hits the target is 42.8 m/s [E].
Copyright © 2003 Nelson
Chapter 1 Kinematics
17
Try This Activity: Graphing Motion with Acceleration
(Page 23)
The required graphs are shown below, in which the position of zero displacement is located where the cart is near the bottom
of the ramp but is not experiencing a push.
PRACTICE
(Pages 23–24)
Understanding Concepts
8. (a) To determine the average acceleration from a velocity-time graph, determine the slope of the line if the acceleration is
constant.
(b) To determine the change in velocity from an acceleration-time graph, determine the area under the line.
9. (a) The motion starts with a westward velocity, but constant eastward acceleration. The motion then slows down to zero
velocity, then accelerates westward with increasing westward velocity. The magnitude of the westward acceleration is
somewhat less than the magnitude of the eastward acceleration.
(b) The motion is southward with northward acceleration slowing down to zero velocity.
(c) The motion is forward with constant acceleration forward. After a period of time, the motion increases to a higher
constant acceleration forward.
(d) The motion starts with northward acceleration then increases its northward acceleration. It starts to slow down
(southward acceleration), and then decreases its southward acceleration to zero.
10. (a)
18
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b)
11.
12. The car’s displacement is the area under the velocity-time graph. It is determined by adding the areas of rectangles and
triangles contained in each time segment. Referring to the figure in the text:
displacement = total area = A4 (0 to 3 s) + A5 (3 s to 5 s) +A6 (5 s to 9 s)
1
(12 m/s [S])(3.0s) = 18 m [S]
2
1
A5 = (12 m/s [S])(2.0s) + (18 m/s [S] − 12 m/s [S])(2.0s) = 30 m [S]
2
1
A6 = (18 m/s [S])(4.0s) + (24 m/s [S] − 18 m/s [S])(4.0s) = 84 m [S]
2
A4 =
displacement = A4 + A5 + A6 = 132 m [S]
The car’s displacement is 132 m [S].
Making Connections
13. (a) The word “idealized” means that the acceleration changes instantaneously from one value to another. In real situations,
changes from one acceleration value to another occur over a finite time interval.
(b) Calculations are much easier if idealized examples are used. For example, to find the change in velocity for an
idealized acceleration graph, we can find the area of a rectangle on the graph. That is much easier than finding the area
under a curved line.
(c)
14. The solution to this question depends on the software, calculator, or planimeter available. Each device is accompanied by
a set of instructions that students can follow to analyze graphs.
Copyright © 2003 Nelson
Chapter 1 Kinematics
19
PRACTICE
(Page 27)
Understanding Concepts
G
2
G G
a ( ∆t )
15. (a) ∆d = vi ∆t +
.
2
G 1 G G
(b) ∆d = (vi + vf ) ∆t .
2
2
2
16. vf = vi + 2a∆d
2
2
2
2
L ?L
 L 
 T  =  T  + 2  2  (L )
   
T 
2
L ?L
L
 T  = T  + 2 T 
   
 
Since the dimensions of each term are the same, the equation is dimensionally correct.
G 1 G G
17. (a) ∆d = (vi + vf ) ∆t
2
G
2 ∆d
∆t = G G
vi + vf
G 1 G G
∆d = (vi + vf ) ∆t
(b)
2
G
G G
2 ∆d
vi + vf =
∆t
G
G  2 ∆d  G
vf = 
−v
 ∆t  i


18. Start with the defining equation for constant acceleration and the equation for displacement in terms of average velocity:
G
G G
G ∆v
a=
∆d = vav ∆t
∆t
G G
G G
G (vi + vf )
G (vf − vi )
∆d =
∆t
a=
2
∆t
(a) To derive the constant acceleration equation in which the final velocity has been eliminated: first solve acceleration
G
equation for vf , then substitute into the equation for displacement.
G G
G (vi + vf )
∆d =
∆t
2
G G G
G (vi + vi + a ∆t )
G G
G vf − vi
∆
=
∆t
d
a=
2
∆t
G
G
G G G
2vi ∆t + a ∆t 2
vf = vi + a ∆t
=
2
G G
1G
∆d = vi ∆t + a ∆t 2
2
20
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) To derive the constant acceleration equation in which the initial velocity has been eliminated: first solve the
G
acceleration equation for vi , then substitute into the equation for displacement.
G G
G (vi + vf )
∆d =
∆t
2
G G
G
G (vf − a ∆t + vf )
∆d =
∆t
G G G
2
vi = vf − a ∆t
G
G
2vf ∆t − a ∆t 2
=
2
G G
1G
∆d = vf ∆t − a ∆t 2
2
G
2
19. a =18 m/s [E]
∆t = 1.6 s
G
vi =73 m/s [W]
G
vf = ?
G G
G v − vi
a= f
∆t
G G G
vf = vi + a ∆t
= 73m/s[W] + 18 m/s 2 [E](1.6s)
G
vf = 44 m/s[W]
The velocity of the birdie is 44 m/s [W].
G
20. vi = 41 m/s [S]
G
vf = 47 m/s [N]
∆t = 1.9 ms = 1.9 × 10–3 s
G
a =?
G G
G v −v
a= f i
∆t
47 m/s[N] − 41m/s[S]
=
1.9 × 10−3 s
G
a = 4.6 × 104 m/s 2 [N]
The acceleration is 4.6 × 104 m/s2 [N].
21. Note: As was described in the text, pages 24 and 25, and applied in Sample Problem 6, page 26, we can omit the vector
G
G
notation when vf 2 or vi 2 terms are involved. However, in the solution shown here as well as the solution for question 24,
we have kept the vector notation in order to show what the final direction is.
G
vi = 0
G
a = 2.3 m/s2 [fwd]
∆t = 3.6 s
G
(a) ∆d = ?
G G
1G
2
∆d = vi ∆t + a ( ∆t )
2
(2.3m/s [fwd ]) (3.6 s)
= 0.0 m +
2
G
∆d = 15 m [fwd]
The sprinter’s displacement is 15 m [fwd].
Copyright © 2003 Nelson
2
2
Chapter 1 Kinematics
21
G
(b) vf = ?
G G
G vf − vi
a=
∆t
G G G
vf = vi + a ∆t
(
)
= 0 m/s + 2.3 m/s 2 [fwd] (3.6 s)
G
vf = 8.3 m/s [fwd]
The sprinter’s final velocity is 8.3 m/s [fwd].
G
22. vi = 7.72 × 107 m/s [E]
G
vf = 2.46 × 107 m/s [E]
G
∆d = 0.478 m [E]
G
(a) a = ?
G
G
G G
vf 2 = vi 2 + 2a ∆d
G
G
G v 2 −v2
a = f Gi
2 ∆d
(2.46 ×10 m/s [E]) − (7.72 ×10 m/s [E])
=
2
7
7
2
2(0.478 m [E])
G
a = −5.60 × 1015 m/s 2 [E]
G
a = 5.60 × 1015 m/s 2 [W]
The electron’s acceleration is 5.60 × 1015 m/s2 [W].
(b) ∆t = ?
G 1 G G
∆d = (vi + vf ) ∆t
2
G
2∆d
t
∆ = G G
vi + vf
=
2 ( 0.478 m [E])
(7.72 ×10 m/s [E]) + (2.46 ×10 m/s [E])
7
7
∆t = 9.39 × 10−9 s
The acceleration occurs over 9.39 × 10–9 s.
Applying Inquiry Skills
23. The experiment can be simple. Besides the book and the desk, the only equipment required is a metre stick and a
stopwatch. Slide the book along the desk at a constant speed for a predetermined distance that is long enough that the time
interval to cover the distance is at least 2.0 s (e.g., slide the book at a constant speed of about 0.50 m/s for 2.0 s). Remove
the pushing force from the book and determine the displacement from that instant to the stopping position. The
acceleration can be found by applying the equation vf 2 = vi 2 + 2a∆d , where vf = 0, vi is the measured value of the speed
while the book is being pushed, and ∆d is the distance the book slides after it is no longer pushed.
Making Connections
G
 1000 m  1 h 
24. vi = 75.0 km/h [N] = (75.0 km/h [N]) 

 = 20.8 m/s [N]
 km  3600 s 
G
a = 4.80 m/s2 [S]
reaction time = ?
22
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
First we must calculate the change in displacement:
G 2 G2
G G
vf = vi + 2a ∆d
G vGf 2 − vGi 2
∆d =
G
2a
=
(0 m/s )2 − ( 20.8 m/s[N])2
2(4.80 m/s 2 [S])
G
∆d = 45.2 m[N]
Calculate reaction distance (distance before stopping) = 48.0 m – 45.2 m = 2.8 m
reaction distance
reaction time =
vi
=
2.8 m
20.8 m/s
reaction time = 0.13s
Thus, the reaction time is 0.13 s.
PRACTICE
(Page 29)
Understanding Concepts
G
25. vi = 25 m/s [E]
G
vf =25 m/s [S]
∆t = 15 s
G
aav = ?
Using components, with +x eastward and +y southward:
∆v x = vfx + ( − vix )
∆v y = vfy + ( − viy )
∆v x = −25 m/s
∆v y = 25 m/s
G
∆v = ∆v x 2 + ∆v y 2
=
( 25 m/s )2 + ( 25 m/s )2
G
∆v = 35.3 m/s
tanθ =
∆v y
∆v x
 25 m/s 
θ = tan −1 

 25 m/s 
θ = 45°
G
So, ∆v = 35m/s [45° S of W ]. Therefore,
G
∆v
G
aav =
∆t
35 m/s [45° S of W ]
=
15 s
G
2
aav = 2.4 m/s [45° S of W ]
Thus, the car’s average acceleration is 2.4 m/s2 [45° S of W].
Copyright © 2003 Nelson
Chapter 1 Kinematics
23
G
26. vi = 6.4 m/s [E]
G
aav = 2.0 m/s2 [S]
∆t = 2.5 s
G
vf = ?
G G
v − vi
G
aav = f
∆t
G G G
vf = vi + aav ∆ t
(
)
= 6.4 m/s [E] + 2.0 m/s 2 [S] ( 2.5s )
G
vf = 6.4 m/s [E] + 5.0 m/s [S]
G
vf = vfx 2 + vfy 2
(6.4 m/s )2 + (5.0 m/s )2
=
G
vf = 8.1m/s
tan θ =
vfy
vfx
 5.0 m/s 
θ = tan −1 

 6.4 m/s 
θ = 38°
The final velocity of the watercraft is 8.1 m/s [38° S of E].
G
27. vi = 26 m/s [22° S of E]
G
vf = 21 m/s [22° N of E]
G
aav = ?
Using +x eastward and +y northward:
∆vx = ( 21m/s ) cos 22° − ( 26 m/s ) cos 22°
∆v y = ( 21m/s ) sin 22° − ( −26 m/s ) sin 22°
∆v y = 17.6 m/s
∆vx = −4.6 m/s
∆v x
∆t
−4.6 m/s
=
2.5 × 10−3 s
= −1.9 ×103 m/s 2
aav,x =
aav,y =
aav,x
aav,y
∆v y
∆t
17.6 m/s
=
2.5 × 10−3 s
= 7.0 ×103 m/s 2
G
aav = aavx 2 + aavy 2
=
(−1.9 ×10 m/s ) + (7.0 ×10 m/s )
3
2 2
3
2 2
G
aav = 7.3 × 103 m/s 2
tan θ =
aavy
aavx
 7.0 × 103 m/s 2 
θ = tan −1 
3
2 
 1.9 × 10 m/s 
θ = 75°
Thus, the average acceleration of the puck is 7.3 × 103 m/s2 [75° N of W].
24
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
28. aav =9.8 m/s2 [down]
∆t = 2.0 s
G
vf =24 m/s [45° below horizontal]
G
vi = ?
G G
v − vi
G
aav = f
∆t
G G G
vi = vf − aav ∆t
(
)
= 24 m/s [45° below the horizontal] − 9.8 m/s 2 [down ] ( 2.0s )
G
vi = 24 m/s [45° below the horizontal] − 19.6 m/s [down ]
vix = ( 24 m/s ) cos 45°
viy = ( −24 m/s ) sin 45° − ( −20 m/s )
vix = 17 m/s
viy = 2.6 m/s
G
vi = vix 2 + viy 2
=
(17 m/s )2 + ( 2.6 m/s )2
G
vi = 17 m/s
tanθ =
viy
vix
 2.6 m/s 
θ = tan −1 

 17 m/s 
θ = 10°
The ball’s initial velocity is 17 m/s [10° above the horizontal].
 82.0 km   1000 m   1 h 
29. vi = 82.0 km/h = 


 = 22.8 m/s
 1 h   1 km   3600 s 
vi = vf = 22.8 m/s
 60 s 
3
∆t = 15 min = 15 min 
 = 9.00 × 10 s
 min 
As stated in the question, +x east and +y north.
∆v
aav,x = x
∆t
22.8 m/s (cos12.7° ) − 22.8 m/s (sin 38.2° )
=
9.00 × 103 s
aav,x = 9.0 × 10 −3 m/s 2
aav,y =
=
∆v y
∆t
−22.8 m/s (sin12.7° ) − 22.8 m/s (cos 38.2° )
9.00 × 103 s
aav,y = −2.5 × 10 −2 m/s2
Thus, the x-component of the average acceleration is 9.0 × 10–3 m/s2 and the y-component is –2.5 × 10–2 m/s2.
Section 1.2 Questions
(Pages 30–31)
Understanding Concepts
1.
2.
Instantaneous acceleration equals average acceleration during motion of constant acceleration.
It is possible to have a northward velocity with westward acceleration if there is a change in direction. For example, if a
truck is initially moving northward at 50 km/h and changes direction to obtain a final velocity of 50 km/h [45° W of N],
the change in velocity and, thus, the acceleration just at the instant of the initial velocity, is westward.
Copyright © 2003 Nelson
Chapter 1 Kinematics
25
3.
G
vi = 1.65 × 103 km/h [W]
G
vf = 1.12 × 103 km/h [W]
∆t = 345 s
G
(a) a = ?
G G
G vf − vi
a=
∆t
1.12 × 103 km/h [W] − 1.65 × 103 km/h [W ]
=
345 s
G
a = 1.54 (km/h)/s [E ]
The average acceleration of the aircraft is 1.54 (km/h)/s [E].
G
 1000 m  1 h 
(b) a = 1.54 (km/h)/s 


 km  3600 s 
G
a = 0.427 m/s 2 [E]
The average acceleration of the aircraft is 0.427 m/s2 [E].
4. (a)
(b) To determine the instantaneous acceleration at t = 2.0 s, calculate the slope of the tangent to the curve indicated on the
graph.
G G
5. (a) Students can determine the data for the velocity-time graph by using the constant acceleration equation ∆d = vav ∆t
(applied at the times indicated), or by drawing the position-time graph and finding the tangents to the curve. The table
below gives the data. The velocity-time graph is a straight line, and its slope indicates the acceleration.
t (s)
G
v (m/s [W])
26
0
0.2
0.4
0.6
0.8
0
2.6
5.2
7.8
10.4
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
(b) a = ?
G
2
G G
a ( ∆t )
∆d = vi ∆t +
2
G
G 2 ∆d
a=
2
( ∆t )
=
G
(where vi = 0)
2(4.16 m [W])
(0.80 s )
2
G
a = 13 m/s 2
6. (a) The motion starts at a specific location at a high velocity with a negative acceleration, reaching zero velocity midway
through the motion. The motion then undergoes increasing velocity in the opposite direction until reaching the initial
position.
(b) This motion is increasing velocity in the negative direction, followed by a decreasing velocity in the same direction,
eventually reaching zero velocity. This is followed by increasing velocity in the positive direction. The magnitudes of
the negative and positive accelerations are equal.
(c) The motion in this graph starts with a constant velocity, then accelerates at a high rate for a short time before slowing
down with negative acceleration at a lower rate to zero velocity.
G
7. vi =26 m/s [E]
G
a = 5.5 m/s2 [W] = −5.5 m/s2 [E]
∆t = 2.6 s
G
vf = ?
G G
G vf − vi
a=
∆t
G G G
vf = vi + a ∆t
= 26 m/s[E] + ( −5.5 m/s 2 [E])(2.6s)
8.
9.
G
vf = 12 m/s[E]
The car’s velocity is 12 m/s [E].
G
a = −9.7 m/s2 [fwd]
∆t = 2.9 s
G
vi = ?
G G
G vf − vi
a=
∆t
G
G
vi = vf − a ∆t
= 0 − ( −9.7 m/s2 [fwd])(2.9 s)
G
vi = 28 m/s[fwd]
The car’s initial speed is 28 m/s.
The data points for the position-time graph can be found by finding the total area on the velocity-time graph up the each
second. The results are shown in the table.
t (s)
G
d (m [W])
Copyright © 2003 Nelson
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
0
2.5
10.0
22.5
37.5
52.5
67.5
78.8
82.5
78.8
67.5
56.2
52.5
Chapter 1 Kinematics
27
The acceleration-time graph is generated by calculating the slopes of the line segments on the velocity-time graph.
10.
G
a = 4.4 m/s2 [fwd]
∆t = 3.4 s
G
vi = 0
G
(a) vf = ?
G G
G v − vi
a= f
∆t
G
G G
vf = vi + a ∆t
(
)
= 0 + 4.4 m/s 2 [fwd ] (3.4 s)
G
vf = 15 m/s [fwd]
The jumper’s final velocity is 15 m/s [fwd].
G
(b) ∆d = ?
G G
1G
2
∆d = vi ∆t + a ( ∆t )
2
1
= 0 + 4.4 m/s 2 [fwd] (3.4 s)2
2
G
∆d = 25 m [fwd]
The jumper’s displacement is 25 m [fwd].
G
11. vi = 0
G
vf = 2.0 × 107 m/s [E]
G
∆d = 0.10 m [E]
G
(a) a = ?
G
G
G G
vf 2 = vi 2 + 2a ∆d
G
G v2
a= f G
2 ∆d
(2.0 × 107 m/s[E])2
=
2(0.10 m [E])
G
a = 2.0 × 1015 m/s 2 [E]
(
)
The acceleration of the electron is 2.0 × 1015 m/s2 [E].
28
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(b) ∆t = ?
G G
G v − vi
a= f
∆t
G G
vf − vi
∆t = G
a
2.0 × 107 m/s [E] − 0
=
2.0 ×1015 m/s 2 [E]
∆t = 1.0 × 10−8 s
The electron takes 1.0 × 10–8 s to reach its final velocity.
G
12. vi = 204 m/s [fwd]
G
vf = 508 m/s [fwd]
∆t = 29.4 s
G
∆d = ?
G 1 G G
∆d = ( vi + vf ) ∆t
2
( 204 m/s [fwd] + 508 m/s [fwd])
=
(29.4 s )
2
G
∆d = 1.05 × 104 m [fwd]
The displacement of the rocket is 1.05 × 104 m [fwd].
G
13. vi = 0
G
vf = 4.2 × 102 m/s [fwd]
G
∆d = 0.56 m [fwd]
G
(a) vav = ?
G G
vi + vf
G
vav =
2
0 + 4.2 ×10 2 m/s [fwd]
=
2
G
2
vav = 2.1 ×10 m/s [fwd]
The average velocity of the bullet is 2.1 × 102 m/s [fwd].
(b) ∆t = ?
G G
∆d = vav ∆t
G
∆d
∆t = G
vav
=
0.56 m [fwd]
2.1× 10 2 m/s [fwd]
∆t = 2.7 × 10−3 s
The uniform acceleration occurs over 2.7 × 10–3 s.
14. (a) After 45 s, the car and the van have the same velocity (from the graph).
(b) Let the subscript V represent the van and the subscript C represent the car. The displacements of the two vehicles are
equal at some time t, and can be found by determining the areas under the lines on the graphs.
G
G
∆d V = Atriangle,V + Arectangle,V
∆d C = Atriangle,C + Arectangle,C
G
G
G
G
= (vV1 )av ∆tV1 + vV2 ∆tV2
= (vC1 )av ∆tC1 + vC2 ∆tC2
 20 m/s + 0 m/s 
=
 60s + ( 20 m/s )(t − 60s )
2


G
∆d V = 600 m+ ( 20 m/s )(t − 60s )
Copyright © 2003 Nelson
 15 m/s + 0 m/s 
=
 30s + (15 m/s )(t − 30s )
2


G
∆d C = 225 m+ (15 m/s )(t − 30s )
Chapter 1 Kinematics
29
G
G
Set ∆d V = ∆dC and solve for t:
600 m+ ( 20 m/s )(t − 60s ) = 225 m+ (15 m/s )(t − 30s )
(20 m/s ) t − (15 m/s ) t = 225 m − (15 m/s )(30 s ) + ( 20 m/s )(60 s ) − 600 m
t=
375 m
5 m/s
t = 75s
Thus, the van V overtakes the car C at 75 s.
G
G
(c) Using t = 75 s, substitute into the equation for ∆d V or ∆d C .
G
G
G
∆d V = (vV1 )av ∆tV1 + vV2 ∆tV2
 20 m/s + 0 m/s 
=
 60s + ( 20 m/s )( 75s − 60s )
2


= 600 m+ 300 m
G
∆d V = 900 m
The displacement from the intersection when V overtakes C is 900 m.
G
15. vA = 4.4 m/s [31° S of E]
G
vB = 7.8 m/s [25° N of E]
∆t = 8.5 s
G
a =?
Using +x east and +y north, we find the components of the velocities and then the accelerations.
vAx = 4.4 m/s (cos 31° )
vBx = 7.8 m/s (cos 25° )
vAx = 3.8 m/s
vBx = 7.1m/s
vAy = −4.4 m/s (sin 31° )
vBy = 7.8 m/s (sin 25° )
vAy = −2.3 m/s
vBy = 3.3m/s
vBx − vAx
∆t
7.1m/s − 3.8 m/s
=
8.5s
ay =
ax =
a x = 0.39 m/s
=
2
vBy − vAy
∆t
3.3 m/s − ( −2.3m/s )
8.5s
a y = 0.66 m/s 2
G
a = ax 2 + a y 2
=
(0.39 m/s ) + (0.66 m/s )
2 2
2 2
G
a = 0.76 m/s 2
tan θ =
ay
ax
 0.66 m/s 2 
θ = tan −1 
2 
 0.39 m/s 
θ = 31°
The bird’s average acceleration is 0.76 m/s2 [31° E of N].
30
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
16. vi = 155 km/h [E]
G
vf = 118 km/h [S]
∆t = 56.5 s
G
a =?
Using components with +x east and +y south:
vfy − viy
v −v
ay =
a x = fx ix
∆t
∆t
0 km/h − 155 km/h
118 km/h − 0 km/h
=
=
56.5s
56.5s
a x = −2.74 (km/h)/s
a y = 2.09 (km/h)/s
G
aav = a x 2 + a y 2
( −2.74 (km/h)/s )2 + ( 2.09 (km/h)/s )2
=
G
aav = 3.45 (km/h)/s
tan θ =
ay
ax
 2.74 (km/h)/s 
θ = tan −1 

 2.09 (km/h)/s 
θ = 52.7°
Thus, the helicopter’s average acceleration is 3.45 (km/h)/s [52.7° W of S].
Applying Inquiry Skills
17. In both cases, the variable most difficult to measure is the time interval for the acceleration. Since the final speed is zero
and the initial speed is given, it remains to find the displacement the object undergoes during the acceleration. Then the
variables can be used in the equation vf 2 = vi 2 + 2a∆d . Estimates of the acceleration will vary if they are just guesses, but
should be fairly close if they are determined by a quick calculation with estimated quantities.
(a) The displacement of the bullet during stopping can be found by measuring the penetration of the bullet plus the
distance the wood moves. Assuming the value is 15 cm [fwd], the acceleration is:
G 2 G2
v −v
G
aav = f G i
2 ∆d
=
0 − (175 m/s [fwd])
2
2(0.15 m [fwd])
G
5
2
aav = −6.8 × 10 m/s [fwd]
The average acceleration of the bullet is –6.8 ¯ 105 m/s2 [fwd].
(b) The stopping distance can be measured by adding the total crunch distance of the car plus any change of position of the
barrels. Assuming the value is 1.5 m [fwd], and changing the initial speed to 24 m/s, the acceleration is:
G 2 G2
vf − vi
G
aav =
G
2 ∆d
=
0 − (24 m/s) [fwd])
2
2(1.5 m [fwd])
G
2
2
aav = −1.9 × 10 m/s [fwd]
The average acceleration of the test car is –1.9 ¯ 102 m/s2 [fwd].
Copyright © 2003 Nelson
Chapter 1 Kinematics
31
Making Connections
18. In the relay, the second, third, and fourth runners have already accelerated to vf before they handoff to the next runner.
Thus, the time is shorter than if you include acceleration time for all runners.
1.3 ACCELERATION DUE TO GRAVITY
PRACTICE
(Pages 32–33)
Understanding Concepts
1.
2.
The skydiver’s velocity is much greater than the diver’s velocity. Air resistance increases with velocity and cannot be
neglected for the skydiver.
The disadvantage is that what might seem logical or reasonable does not agree with what actually happens. One reason for
this is some events occur too rapidly for our senses to be able to observe slight differences. Aristotle’s reasoning that
heavy objects fell faster than lighter ones provides an example of the disadvantage of not using experimentation to
determine the dependency of one variable on another, in this case the dependency of the acceleration of a falling body on
the mass of the body.
Applying Inquiry Skills
3.
The experimental setup would require a vacuum chamber in which the coin and the feather are released simultaneously.
(This device is available commercially from scientific supply companies.)
Making Connections
4.
Since there is no atmosphere on the Moon, falling objects do not experience air resistance.
Case Study: Predicting Earthquake Accelerations
(Page 34)
(a) The map stretches from Northern California (40° north latitude) to a few kilometres north of Vancouver (50° north
latitude), and from the middle of Vancouver Island (125° west longitude) to just east of Trail, B.C. (about 117° west
longitude). The regions affected most severely, as depicted by deep reds, lie near the west coast of the continent,
especially in Northern California and Southern Oregon, as well as areas fairly close to Vancouver and Victoria. Areas
affected moderately, as depicted by yellows, stretch inland somewhat in British Columbia, Washington, and Oregon, and
a long way in California. Areas affected slightly, as depicted by blue-greens, are in the northern and eastern parts of
British Columbia, the eastern parts of Washington and Oregon, and the state of Idaho. There are no areas unaffected, as
depicted by white.
(b) Students have to combine the colours on the map with the colours in the legend, and they have to compare the contours
and locations on the map to a conventional atlas of the same region.
PRACTICE
(Page 35)
Understanding Concepts
5.
32
G
vi = 0
G
a = 9.80 m/s2 [down]
G
(a) ∆d = 5.00 m
G
vf = ?
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson