Physics 11 Assignment
Chapter 3
Physicist: KEY
1. Define the following terms:
 Vector - a physical quantity that has a magnitude and a direction
 Scalar - a physical quantity that has only a magnitude or size
 Vector diagram - a diagram, with a coordinate system, in which all quantities are
represented by vectors
 Resultant vector – the sum or difference of two or more vectors
 Vector components - parts of a vector that lie on the axes of a coordinate system, into
which a vector can be resolved
 Vector resolution – the process by which a vector is broken up into its perpendicular
components
 Relative velocity - a velocity relative to a frame of reference, such as an air current, that
itself is moving with a velocity relative to another frame of reference, such as the ground
2. Sketch appropriate position-time and acceleration-time graphs for the velocity-time graph below:
a)
b)
3. A car accelerates from 12 m/s to 25 m/s in 6.0 s. What is its acceleration? How far did it travel in
this time? Assume constant acceleration.
We find the acceleration from
v f  vi  at
25 m/s = 12 m/s + a (6.0 s)
25 m/s - 12 m/s
a
 2.17 m/s2
6.0s
We find the distance traveled from
1
d  (v f  vi )t
2
1
d  (25 m/s + 12 m/s)(6.0s) = 111 m
2
4. A car slows down from 20.0 m/s to rest in a distance of 85.0 m. What is its constant acceleration?
We find the acceleration (assumed constant) from
 2 2

v f  vi  2a d
 2 2
 v f  vi
(0m / s) 2  (20m / s) 2

a 

 2.35m / s 2
2(85m)
2d
5. A person driving her car at 50 km/h approaches an intersection just as the traffic light turns yellow.
She knows that the yellow light last only 2.0 s before turning to red, and she is 30.0 m away from
the near side of the intersection. Her car’s maximum deceleration is -6.0 m/s2, whereas it can
accelerate from 50.0 km/h to 70.0 km/h in 6.0 s. Should she try to stop, or should she make a run
for it? Ignore the length of her car and her reaction time.
We use a coordinate system with the origin where the initial action takes place, as shown
in the diagram. The initial speed is (50 km/h)/(3.6 km/h) = 13.89 m/s.
If she decides to stop, we find the minimum stopping distance from
v f 2  vi 2  2ad
d 
v f 2  vi 2
2a
(0 m/s) 2 - (13.89 m/s) 2

 16.08 m
2( 6.0 m/s2 )
Because this is less than 30.0 m, the distance to the intersection, she can safely stop in
time.
If she decides to increase her speed, we find the acceleration to go from 50 km/h to 70
km/h (19.44 m/s):
v f  vi  at
a 
v f  vi
t

(19.44 m/s) - (13.89 m/s)
 0.926 m/s2
6.0 s
We find her location when the light turns red from
1
d  vi t  at 2
2
1
d  (13.89 m/s)(2.0 s)+ (0.926 m/s2 )(2.0 s) 2 = 29.6 m
2
Because this is less than 30.0 m, she is not in the intersection (but almost), but moving at
high speed. Given that the length of her car and her reaction time were not considered, the
safer action would be to STOP!
6. A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.50 s. How high
is the cliff?
We use a coordinate system with the origin at the top of the cliff. To find the height of the
cliff,
d  vi t 
1 2
gt
2
1
d  (0 m/s)(3.50 s)+ (-9.81 m/s2 )(3.50 s)2 = -60.1 m
2
Note: the “-“ sign in the answer indicates that the stone fell DOWN.
7. A foul ball is hit straight up into the air with a speed of 25 m/s. (a) How high does it go? (b) How
long is it in the air?
(a)
v f 2  vi 2  2ad
d 
v f 2  vi 2
2a

(0 m/s) 2 - (25.0 m/s) 2
 31.9 m
2( 9.81 m/s2 )
(b)
d  vi t 
1 2
gt
2
1
0 m  (25 m/s)(t )+ (-9.81 m/s 2 )(t ) 2
2
Factor out "t" from the right side of the equation to get
1
0 m  t (25 m/s + (-9.81 m/s 2 )t )
2
Solve for "t"
t = 0 or 25 m/s +
1
(-9.81 m/s 2 )t = 0
2
 t = 5.1 s
8. Two vectors are added together, and their vector sum is zero. What can you say about the
magnitude and direction of the two initial vectors?
The magnitude of the two vectors is the same, but the direction is opposite (i.e. 180o). For
example, A = 2 km [West] is added to B = 2 km [East]. A + B = 0 km.
9. Can the displacement vector for an object moving in two dimensions ever be longer than the
length of path traveled by the object over the same time interval? Can it ever be less? Discuss.
The displacement can be thought of as the “straight line” path from the initial location to
the final location. The length of path will always be greater than or equal to the
displacement, because the displacement is the shortest distance between the two
locations. Thus the displacement can never be longer than the length of path, but it can be
less. For any path that is not a single straight line segment, the length of path will be longer
than the displacement.
10. Two vectors have magnitudes V1= 3.5 km and V2= 4.0 km. What are the maximum and minimum
magnitudes of their vector sum?
If the two vectors are in the same direction, the magnitude of their sum will be a maximum,
and will be 7.5 km. If the two vectors are in the opposite direction, the magnitude of their
sum will be a minimum, and will be 0.5 km. If the two vectors are oriented in any other
configuration, the magnitude of their sum will be between 0.5 km and 7.5 km.
11. Can two vectors, of unequal magnitude, ever add up to give a resultant with a magnitude of zero?
Can three unequal vectors? Under what conditions?
Two vectors of unequal magnitude can never add to give the
zero vector. However, three vectors of unequal magnitude can
add to give the zero vector. If their geometric sum using the
tail-to-tip method gives a closed triangle, then the vector sum
will be zero. See the diagram, in which
.
12. Can the magnitude of a vector ever (a) equal, or (b) be less than, one of its components?
(a) The magnitude of a vector can equal the length of one of its components if the other
components of the vector are all 0; i.e. if the vector lies along one of the coordinate axes.
(b) The magnitude of a vector can never be less than one of its components, because each
component contributes a positive amount to the overall magnitude, through the
Pythagorean relationship. The square root of a sum of squares is never less than the
absolute value of any individual term.
13. If you are riding on a train that speeds past another train moving in the same direction on an
adjacent track, it appears that the slower moving train is moving backwards. Why?
Your reference frame is that of the train you are riding. If you are traveling with a relatively
constant velocity (not over a hill or around a curve or drastically changing speed), then you
will interpret your reference frame as being at rest. Since you are moving forward faster
than the other train, the other train is moving backwards relative to you. Seeing the other
train go past your window from front to rear makes it look like the other train is going
backwards. This is similar to passing a semi-truck on the highway – out of a passenger
window, it looks like the truck is going backwards.
14. Two rowers, who can row at the same speed in still water, set off across a river at the same time.
One heads straight across and is pulled downstream somewhat by the current. The other one
heads upstream at an angle so as to arrive at a point opposite the starting point. Which rower
reaches the opposite side first?
Both rowers need to cover the same "cross river" distance. The rower with the greatest
speed in the "cross river" direction will be the one that reaches the other side first. The
current has no bearing on the problem because the current doesn't help either of the boats
move across the river. Thus the rower heading straight across will reach the other side
first. All of his rowing effort has gone into crossing the river. For the upstream rower, some
of his rowing effort goes into battling the current, and so his "cross river" speed will be
only a fraction of his rowing speed.
15. A delivery truck travels 14 blocks north, 16 blocks east, and 26 blocks south. What is its final
displacement from the origin? Assume the blocks are equal in length.
We choose the north and east coordinate system shown. For the components of the
resultant, we have:
East: Rx = D2 = 16 blocks
North: Ry = D1 – D3 = 14 blocks – 26 blocks = -12 blocks
We find the resultant displacement from:

2
2
 12 
R  Rx  Ry
  tan  `1  
 16 

2
2
R  16   12
  37 0 South of East

R  20 blocks
16. If Vx =18.80 units and Vy = -16.40 units, determine the magnitude and direction of V.
We find the vector from:

2
2
V  V x  Vy

2
2
V  18.80   16.40

V  25.0
  16.40 

 18.80 
  41.10 below the x - axis
  tan `1 
17. For the vectors given in the figure below, determine (a) A – C, (b) B – A, (c) A – B + C, (d) A + B C, and (e) B - 2A.
(a)
(b)
(c) Use vector resolution to solve for the resultant vector, R. Note: R = A – B + C = A + (-B) +
C

2
2
  48.97 
R

R

R
  tan  `1 

x
y
X
Y
80.64 

Vector

component Component
2
2

R  80.64   48.97
  31.3 0 South of East
o
o
66.0cos28
66.0sin28
A


R  94.4 units
40.0cos56o -40.0sin56o
 B
0
-46.8
C

80.64
-48.97
R
(d) R = A + B - C = A + B + (-C)
X
Y
Vector
component Component

66.0cos28o 66.0sin28o
A

40.0sin56o
B
40.0cos56o

0
46.8
C

35.91
110.95
R
 110.95 

 35.91 
  72.10 North of East
  tan  `1 

2
2
R  Rx  Ry

2
2
R  35.91  110.95

R  117 units
(e) R = B -2A = B + (-2A)
Vector

B

 2A

R
X component
-40.0cos56o
2(66.0cos28o)
-138.92
Y
Component
40.0sin56o
-2(66.0sin28o)

2
2
R  Rx  Ry

2
R   138.92 2   28.81

R  142 units
-28.81
 28.81 

 138.92 
  11.7 0 South of West
  tan `1 
18. A boat can travel 2.30 m/s in still water. (a.) If the boat points its prow directly
across a stream whose current is 1.20 m/s, what is the velocity (magnitude and
direction) of the boat relative to the shore? (b) What will be the position of the boat,
relative to its point of origin, after 3.00 s?
(a) If Vbs is the velocity of the boat with respect to the shore, V bw the velocity of the boat
with respect to the water and Vws the velocity of the water with respect to the shore,
then:
Vbs = Vbw + Vws, as shown in the diagram. From the diagram, we get:

2
2
 2.30 m/s 
Vbs  Vbw  Vws
  tan  `1 

 1.20 m/s 

2
2
V  2.30 m/s  1.20 m/s
  62.4 0 from the shore

V  2.59 m/s
(b) Because the boat will move with constant velocity, the displacement will be:
d  Vbs  t  (2.59 m/s)(3.00 s)  7.77 m at 62.4 o to the shore
19. An airplane is heading due south at a speed of 500 km/h. A wind begins blowing from the
southwest at a speed of 100 km/h. What is the velocity (magnitude and direction) of the plane
relative to the ground? (b) How far off course will it be after 10 min if the pilot takes no corrective
action?
If Vpa is the velocity of the airplane with respect to the air, Vpg the velocity of the airplane with
respect to the ground, and Vag the velocity of the air with respect to the ground, then:
Vpg = Vpa + Vag, as shown in the diagram.
(a)

2
2
V pg  V pa  Vag  2V pa Vag cos 45 o

2
2
V  500 km/h  100 km/h  2 500 km/h 100 km/h cos 45 o

V  435 km/h
 100km / h  sin45 o
435 km/h

  sin -1 

  9.36 0 east of south

(b) Because the pilot is expecting to move south, we find the
easterly distance from this line by:
d  Vag  t  (100km / h)(cos45 o )(10 minutes)  (60 minute/hour)  11.8 km