Chapter 3 Motion in a Plane
Vectors and scalars
Vectors have direction as well as magnitude. They are represented by arrows. The arrow points
in the direction of the vector and its length is related to the vector’s magnitude.
Scalars only have magnitude.
We write A = B if the vectors have the same magnitude and point in the same direction.
Scalars can have magnitude, algebraic sign, and units. Adding scalars is very familiar. You add
10 grams to 15 grams and get 25 grams. You have $20 and give $5 to friend and you have $15
remaining.
Vector addition is different since vectors have direction as well as magnitude. How do we add
vectors?
We already know how to add vectors in one dimension (along the x-axis for example).
A
B
A+B
What happened? The vector B is positioned so that the tail of B is positioned at the head of A.
The vector sum is drawn from the tail of A to the head of B.
If A is 8 m long and B is 10 m long, the magnitude of A+B is 18 m. What if B is reversed?
B
A
A+B
What is the magnitude of A+B?
Here we see a hint of the problem. Vectors do not add like scalars.
How do we add vectors that do not point along the same direction?
3-1
1. Draw the first vector in the correct direction and with the appropriate magnitude.
2. Draw the second vector with the correct direction and magnitude so that its tail is placed
at the head of the first vector.
3. If there is a third vector, draw it with the correct direction and magnitude to that its tail is
placed at the head of the second vector.
4. When finished with all the vectors, find the vector sum by drawing a vector that starts at
the tail of the first vector and ends at the head of the last vector.
How do we subtract vectors? Use A−B = A+(−B). What is a reasonable definition for –B? The
negative of a vector has the same magnitude as the original vector but points in the opposite
direction.
The idea of vectors is built from the idea of displacement. In the diagram above, imagine that
you are in a forest. A is your walk to a tree and B is your walk from the first tree to a friend you
see across the forest. A+B is your net displacement from your starting point.
Another example:
This procedure is called the graphical addition of vectors. You need to understand this
procedure. However, it is too slow and imprecise to be used in solving problems.
3-2
We add vectors by taking their components. The process is summarized in this figure.
We are adding two vectors that are not collinear. We replace each vector with two vectors
(called its components). We then add like components together, giving the components of the
vector sum. What happens next?

C

Cy

Cx

We add Cx and Cy to find C. Since the x- and y-axes are perpendicular, we can find the
magnitude of C from the Pythagorean theorem,
C  Cx  C y
2
2
The direction is normally measured counterclockwise from the +x-axis. For this vector in the 2nd
quadrant, first find 
  arctan
Cy
Cx
and then subtract from 180º. (Why?)
How do we find the components of a vector? As an example suppose A has magnitude 20 N and
it points at 40º. As the following diagram shows, we are dealing with a right triangle. To find
the components we need to use trigonometry. Recall,
cos  
adjacent
hypotenuse
sin  
opposite
hypotenuse
3-3
tan 
opposite
adjacent

A

Ay
=40o

Ax
Using the definitions of cosine and sine,
cos  
A
adjacent
 x
hypotenuse A
Ax  A cos   (20 N) cos 40   15.3 N
sin  
Ay
opposite

hypotenuse A
Ay  A sin   (20 N) sin 40   12.9 N
Why is Ax > Ay? When are they equal?
Suppose we had this picture. What would you do?

Ay

A
=50o

Ax
Ax  (20 N) sin 50   15.3 N
Ay  (20 N) cos 50   12.9 N
Usually we have cosine associated with x-components and sine associated with y-components,
but not always. You have to look at the diagram. (A very common remark for this semester!)
Problem-Solving Strategy: Finding the x- and y-components of a Vector from its
Magnitude and Direction (page 60)
1. Draw a right triangle with the vector as the hypotenuse and the other two sides parallel to
the x- and y-axes.
2. Determine one of the unknown angles in the triangle.
3. Use trigonometric functions to find the magnitudes of the components. Make sure your
calculator is in “degree mode” to evaluate trigonometric functions of angles in degrees
and in “radian mode” for angles in radians.
4. Determine the correct algebraic sign for each component.
3-4
Problem-Solving Strategy: Finding the Magnitude and Direction of a Vector A from its xand y-components (page 60)
1. Sketch the vector on a set of x- and y-axes in the correct quadrant, according to the signs
of the components.
2. Draw a right triangle with the vector as the hypotenuse and the other two sides parallel to
the x- and y-axes.
3. In the right triangle, choose which of the unknown angles you want to determine.
4. Use the inverse tangent function to find the angle. The lengths of the sides of the triangle
represent |Ax| and |Ay|. If  is opposite the side parallel the side perpendicular to the xaxis, then tan  = opposite/adjacent = |Ax/Ay|. If  is opposite the side parallel the side
perpendicular to the y-axis, then tan  = opposite/adjacent = |Ay/Ax|. If your calculator is
in “degree mode,” then the result of the inverse tangent will be in degrees. [In general,
the inverse tangent has has two possible values between 0 and 360º because tan  = tan
( + 180º). However, when the inverse tangent is used to find one of the angles in a right
triangle, the result can never be greater than 90º, so the value the calculator returns is the
one you want.
5. Interpret the angle: specify whether it is the angle below the horizontal, or the angle west
of south, or the angle clockwise from the negative y-axis, etc.
6. Use the Pythagorean theorem to find the magnitude of the vector.
A  Ax  Ay
2
2
Problem-Solving Strategy: Adding Vectors Using Components (page 61)
1. Find the x- and y-components of each vector to be added.
2. Add the x-components (with their algebraic signs) of the vectors to find the x-component
of the sum. (If the signs are not correct, the sum will not be correct.
3. Add the y-components (with their algebraic signs) of the vectors to find the y-component
of the sum.
4. If necessary, use the x- and y-components of the sum to find the magnitude and direction
of the sum.
Even when using the component method to add vectors, the graphical method is an important
first step. Graphical addition gives you a mental picture of what is going on.
A problem can be made easier to solve with a good choice of axes. Common choices are
 x-axis horizontal and y-axis vertical, when the vectors all lie in the vertical plane;
 x-axis east and y-axis north, when the vectors lie in a horizontal plane; and
 x-axis parallel to an inclined surface and y-axis perpendicular to it.
Read the section on unit vectors on pages 62-63. We will not use the unit vector approach, but
you may be familiar with it.
3-5
We do not deal with vectors, we deal with their components.
Here is an algorithm for adding vectors. The diagram is Figure 3.9 on page 61.
  
C AB
Given A, A
and B, B
Bx
Ax
By
B
A
Ay
Find components*
Ax = A cos A, Ay = A sin A
Bx = B cos B, By = B sin B
C
Add like components
Cx = Ax + Bx ,
Cy = Ay + By
Cx
Bx
Ax
By
Return to magnitude and
direction format
C  Cx  C y
2
 C  tan 1
Cy
Cx
Ay
2
Cy
**
C
Cy
C
Cx
*Be careful with the angles given. The equations hold for angles measured counterclockwise
from the +x-axis.
**Be careful with tan-1 function on your calculator. If the x-component is negative, add 180o to
the value found by your calculator.
3-6
Now let’s use the concept of vectors to extend the kinematical variables to more dimensions.
Average velocity is the displacement over the time,


r
v av 
t
Instantaneous velocity is


r
v  lim
t 0 t
The velocity is tangent to the path of the particle.
The average acceleration is


v
a av 
t
Instantaneous acceleration is


v
a  lim
t 0 t
“For straight-line motion the acceleration is always along the same line as the velocity. For
motion in two dimensions, the acceleration vector can make any angle with the velocity vector
because the velocity vector and change in magnitude, in direction, or both. The direction of the
acceleration is the direction of the change in velocity v during a very short time.”
The above definitions look good, but they are not useful. We call these formal definitions. They
are not used in solving problems. Instead we need a set of equations for the x- and ycomponents. The basic rule is
WE DO NOT DEAL WITH VECTORS. WE DEAL WITH THEIR COMPONENTS.
3-7
For the velocity, we have
v x ,av 
x
t
v y ,av 
y
t
with similar definitions for the other parameters. (See pages 64-65.)
We can now generalize the equations at the top of these notes to two dimensions. “It is generally
easiest to choose the axes so that the acceleration has only one non-zero component.” We
choose the y-axis along the direction of acceleration. This means ax = 0. The first equation,
v x  v fx  vix  a x t
becomes two equations:
v x  0
v y  v fy  viy  a y t .
and
x  12 (v fx  vix )t
becomes two equations as well:
x  v x t
y  12 (v fy  viy )t .
and
x  vix t  12 a x (t ) 2
becomes
x  v x t
y  viy t  12 a y (t ) 2 .
and
v 2fx  vix2  2a x x
becomes
v 2fx  vix2  0
Summary (see page 67):
x-axis : ax = 0
and
v 2fy  viy2  2a y y .
v x  0
y-axis: constant ay
v y  v fy  viy  a y t
Equation
(3-19)
x  v x t
y  12 (v fy  viy )t
(3-20)
y  viy t  12 a y (t ) 2
(3-21)
v 2fy  viy2  2a y y
(3-22)
3-8
Projectiles are a good example of this type of motion. Here ay = −g.
This motion is simultaneous constant velocity in the x-direction with constant acceleration in the
y-direction.
3-9
A golf ball is hit from the ground at 35 m/s at an angle of 55º. The ground is level.
1. How long is the ball in the air?
2. What is the maximum height of the ball?
3. How far from the launching point does the ball hit the ground?
4. What is the ball’s position after 2 seconds. When does it reach this height again?
5. When is the ball 20 m above the ground?
In the plot below, all units are in meters.
45
40
35
30
25
20
15
10
5
0
0
20
40
60
80
100
120

vi

v iy
 = 55o

v ix
Solution:
1. When the ball hits the ground, y = 0.
y  viy t  12 a y (t ) 2
0  t (viy  12 a y t )
Since the only way a product can equal zero is when one of the factors equals zero,
t  0
or
(viy  12 a y t )  0
3-10
The first condition tells us that the golf ball starts from the ground. The second gives us
the time of flight, tf
(viy  12 a y t )  0
1
2
a y t  viy
t f  
2viy
ay
This is more useful if we use ay = −g and viy = vi sin  (see the diagram below the
trajectory plot).
t f  

2viy
ay
2vi sin 
g
2(35 m/s ) sin 55

9.8 m/s 2
 5.85 s
What angle maximizes the time of flight? The angle that maximizes sin . The largest
sine can be is 1 and that occurs at 90º. Hit the ball straight up!
2. At the maximum height, vfy = 0.
v fy  viy  a y t
0  vi sin    gth
th 
vi sin 
g
But this is ½ the time of flight. When the ball is shot over level ground half of the time
the ball is going up, the other half of the time it is going down. It takes half the total time
to reach the highest point.
The height at this time is
y  viy t  12 a y (t ) 2
h  vi sin th  12 g (th ) 2
 v sin   1  vi sin  
  2 g 

 vi sin   i
 g 
 g 
3-11
2
vi sin 2  1 vi sin 2 
 g
g
2
g2
2
h
2
vi sin 2 
2g
2

(35 m/s ) 2 sin 2 55
2(9.8 m/s 2 )
 41.9 m/s

What angle maximizes the height? We need to find the maximum of sin2 . The
maximum of sin2  occurs at the maximum of sin , which again is 90º. Hit it straight up.
3. The ball hits the ground when t = tf. The horizontal position where it hits the ground is
called the range (R).
x  vix t
R  vi cos t f
 vi cos 
2vi sin 
g
v 2 sin  cos 
 i
g
2
This can be rewritten using the identity sin 2 = 2sin  cos ,
vi sin 2
g
2
R
(35 m/s ) 2 sin[2(55 )]
9.8 m/s 2
 117 m

What angle maximizes the range? This time we want to find the maximum of sin 2.
The maximum of sine occurs at 90º. This time 2 = 90º or  = 45º. The angle required
for maximum range over level ground is 45º. Also the range is symmetric about 45º. For
some angle ,
sin[2(45   )]  sin(90  2 )  cos 2
and
sin[2(45   )]  sin(90  2 )  cos(2 )  cos 2
3-12
The range for 1 = 45º +  is the same for 2 = 45º − . Another way is to say if
1  2  (45   )  (45   )  90
the ranges are the same!
4. Where is the ball at 2 seconds? Its horizontal position is found using
x  vix t
 vi cos t f
 (35 m/s ) cos 55 2 s 
 40.2 m
Its vertical position is
y  viy t  12 a y (t ) 2
 vi sin t  12 g (t ) 2
 (35 m/s ) sin 55 (2 s)  12 (9.8 m/s 2 )(2 s) 2
 37.7 m
When is it at this height again? There are many ways to find this. First, note that the
time of flight of the ball is 5.85 s from part 1. Since the trajectory is symmetric, if it
reaches this height 2 s after launch, it will reach it again 2 s before it lands,
t  5.85 s  2.00 s  3.85 s
3-13
Another way is to use the symmetry of the y-component of the velocity. The ycomponents of the velocities at the same heights have the same magnitudes but opposite
signs. At 2 s
v fy  viy  a y t
v fy  vi sin   gt
 (35 m/s ) sin 55  (9.8 m/s 2 )(2 s)
 9.07 m/s
When is the speed −9.07 m/s?
v fy  viy  a y t
v fy  vi sin    gt
t 
v fy  vi sin 
g
 9.07 m/s  (35 m/s) sin 55
 9.8 m/s 2
 3.85 s

Finally, the worst way is to just solve for t using the quadratic formula,
y  viy t  12 a y (t ) 2
 vi sin t  12 g (t ) 2
37.7 m  (35 m/s ) sin 55 t  12 (9.8 m/s 2 )(t ) 2
1
2
(9.8 m/s 2 )(t ) 2  (35 m/s ) sin 55 t  37.7 m  0
(4.9 m/s 2 )(t ) 2  (28.67m/s )t  37.7 m  0
t 
 b  b 2  4ac
2a
 (28.67)  (28.67) 2  4(4.9)(37.7)
2(4.9)
28.67  9.11

9.8
 2.00 s, 3.86 s

The first answer is then it reaches 37.7 m while ascending (we knew it would be 2 s), the
second is when it reaches 37.7 m while descending. (I quit writing units in the quadratic
because it makes the equation even more unwieldy.)
3-14
5. To find when the ball reaches 20 m we use the quadratic equation again,
y  viy t  12 a y (t ) 2
 vi sin t  12 g (t ) 2
20 m  (35 m/s ) sin 55 t  12 (9.8 m/s 2 )(t ) 2
1
2
(9.8 m/s 2 )(t ) 2  (35 m/s ) sin 55 t  20 m  0
(4.9 m/s 2 )(t ) 2  (28.67m/s )t  20 m  0
t 
 b  b 2  4ac
2a
 (28.67)  (28.67) 2  4(4.9)(20)
2(4.9)
28.67  20.74

9.8
 0.81s, 5.04 s

Notice that the sum of these times is 5.85 s, the time of flight.
Summary: Derived equations for a projectile launched from level ground with initial velocity
vi at an angle  above the ground:
Time of flight
Time to height
2vi sin 
g
v sin  1
th  i
 2 t f
g
t f 
Maximum height
vi sin 2 
h
2g
Range
v sin 2
R i
g
2
2
If the ground is not level, for example throwing a ball from the top of a building, these
equations will not apply (unless the initial and final heights are the same)!
The projectile travels in a parabolic path as long as we neglect air resistance. The motion is
symmetric about the maximum height (the vertex of the parabola).
Problem: A ball is thrown horizontally from a 30 m tall tower. The ball hits the ground 50 m
from the base of the tower. What is the speed of the ball when it is released?
Treat each component separately. Find the time it takes for the ball to hit the ground and use that
to find the initial velocity.
3-15
y
vi
x
y
x
The time it takes to hit the ground is found from the equation for motion in the y-direction.
y  viy t  12 a y (t ) 2
 0
t 

1
2
 g (t ) 2
 2y
g
 2 30 m 
9. 8 m / s 2
 2.47 s
The ball is 50 m from the base.
 x  v x t
x
t
50 m

2.47 s
 20.2 m/s
vx 
Relative velocity is a great example of adding vectors.
Have you ever had this happen to you? While sitting in your car at a red traffic light, the car
beside you slowly drifts forward. You mash on the brake to stop your car from rolling
backwards, but your car is not moving.
3-16
Within your environment, there is no way to distinguish between your car moving backwards
and the car besides you moving forward. The velocity is relative. We need a reference frame
(the traffic light, for example) to define who is moving.
The train moves at 10 m/s and Wanda can walk at 1 m/s. How fast will Greg see Wanda walk?
Wanda’s velocity relative to Greg is the sum of the velocity of the Wanda relative to the train
plus the velocity of train relative to Greg.



vWG  vWT  vTG
Notice the order of the subscripts. We have the Ts cancelling from the two terms on the right.
This equation will always hold, but how do we use it? What is our rule about vectors?
WE DO NOT DEAL WITH VECTORS. WE DEAL WITH THEIR COMPONENTS.
Take the x-component:
vWGx  vWTx  vTGx
 (1 m/s)  (10 m/s)
 11 m/s
Greg sees Wanda walking to the right at 11 m/s. What happens when she walks back to her seat?
vWGx  vWTx  vTGx
 (1m/s)  (10 m/s)
 9 m/s
According to Greg, Wanda is walking at 9 m/s to the right.
Hopefully, this is pretty easy. But what about this?
3-17
From Example 3.11. Jack wants to row directly across
the river from the east shore to a point on the west shore.
The current 0.60 m/s and Jack can row at 0.90 m/s. What
direction must he point the boat and what is his velocity
across the river?
The velocity of the rowboat relative to the shore is equal
to the velocity of the rowboat relative to the water plus
the velocity of the water relative to the shore.



v RS  v RW  vWS
The rowboat is to head directly to the west.
Take components.
vRSx  vRWx  vWSx
and
vRSy  vRWy  vWSy
The diagram is the key to solving relative velocity problems. For the x-component,
vRSx  vRWx  vWSx
vRS  vRW cos   0
 vRW cos 
The y-component,
vRSy  vRWy  vWSy
0  vRW sin   vWS
vWS  vRW sin 
Our unknowns are  and vRS. From the y-component equation,
3-18
vWS  v RW sin 
sin  
vWS 0.6 m/s

 0.667
v RW 0.9 m/s
  41.8 
From the x-component equation.
vRS  vRW cos 
 (0.90 m/s) cos 41.8
 0.67 m/s
The boat must point 41.8º N of W upstream. Its speed across the water is 0.67 m/s.
3-19