Chapter 3 Assignment – KEY
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
Questions - Page 70
2. 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.
3. 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.
6. 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.
7. 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
.
8. (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.
14. 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.
15. 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.
Problems – Pages 70-75
2. 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
R  Rx  R y

2
2
R  16   12

R  20 blocks
3.
 12 

 16 
  37 0 South of East
  tan `1 
4. We find the vector from:

2
2
V  Vx  V y

2
2
V  18.80   16.40

V  25.0
  16.40 

 18.80 
  41.10 below the x - axis
  tan `1 
7. Because the vectors are parallel, the direction can be indicated by
the sign.
  
C  A  B  8.31  (5.55)  2.76 in the  x direction
  
C  A  B  8.31  (5.55)  13.86 in the  x direction
  
C  B  A  5.55  (8.31)  13.86 in the  x direction
14. (a) Use vector resolution to solve for the resultant vector, R. Note: R = A – B + C = A + (-B) + C
Vector X component Y Component

A
66.0cos28o
66.0sin28o

B
40.0cos56o
-40.0sin56o

C
0
-46.8

R
80.64
-48.97

2
2
R  Rx  R y

2
2
R  80.64   48.97

R  94.4 units
  48.97 

 80.64 
  31.30 South of East
  tan  `1 
(b) R = A + B - C = A + B + (-C)
Vector
X
component
Y
Component

A
66.0cos28o
66.0sin28o

B
-40.0cos56o
40.0sin56o

C
0
46.8

R
35.91
110.95

2
2
R  Rx  Ry

2
2
R  35.91  110.95

R  117 units
 110.95 
  tan  `1 

 35.91 
  72.10 North of East
(c) R = B -2A = B + (-2A)
Vector
X component
Y Component

B
-40.0cos56o
40.0sin56o

 2A
-2(66.0cos28o)
-2(66.0sin28o)

R
-138.92
-28.81

2
2
R  Rx  Ry

2
R   138.92 2   28.81

R  142 units
 28.81 
  tan  `1 

 138.92 
  11.7 0 South of West
Note: this question could have been solved using Cosine and Sine Laws, given that α = 28o
and β = 56o. Therefore, in the vector diagram, the angle between vectors B and -2A is 180o –
(28o + 56o) = 96o.
15. (a) R = C – A – B
Vector X component Y Component

A
-66.0cos28o
-66.0sin28o

B
40.0cos56o
-40.0sin56o

C
0
-46.8

R
-35.91
-110.95

2
2
R  Rx  Ry

2
2
R   35.91   110.95

R  117 units
 110.95 
  tan  `1 

 35.91 
  72.10 South of West
(b) R = 2A – 3B + 2C
Vector
X component
Y Component

2A
2(66.0cos28o)
2(66.0sin28o)

 3B
-3(-40.0cos56o) -3(40.0sin56o)

C
0
2(-46.8)

R
183.65
-131.11

2
2
R  Rx  Ry

2
2
R  183.65   131.11

R  226 units
 131.11 
  tan  `1 

 183.65 
  35.5 0 South of East
43. (a) If Vbs is the velocity of the boat with respect to the shore, Vbw 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
Vbs  Vbw  Vws

2
V  2.30 m/s  1.20 m/s

V  2.59 m/s
 2.30 m/s 

 1.20 m/s 
  62.4 0 from the shore
  tan `1 
2
(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
45. 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
Vpg  Vpa  Vag  2 Vpa 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
  sin -1 
435 km/h


  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/hou r)  11.8 km