3-1

B A

A+B

B

A A+B

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).

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?

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?

3-2

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.

3-3

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?

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,

22

yx CCC

The direction is normally measured counterclockwise from the +x-axis. For this vector in the 2nd

quadrant, first find

x

y

C

Carctan

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,

adjacent

oppositetan

hypotenuse

oppositesin

hypotenuse

adjacentcos

C

yC

xC

3-4

Using the definitions of cosine and sine,

N3.1540cos)N20(cos

hypotenuse

adjacentcos

AA

A

A

x

x

N9.1240sin)N20(sin

hypotenuse

oppositesin

AA

A

A

y

y

Why is Ax > Ay? When are they equal?

Suppose we had this picture. What would you do?

N9.1250cos)N20(

N3.1550sin)N20(

y

x

A

A

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.

A

xA

=50o

=40o

xA

yA

A

yA

3-5

Problem-Solving Strategy: Finding the Magnitude and Direction of a Vector A from its x-

and 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 x-

axis, 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.

22

yx AAA

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.

3-6

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.

BAC

*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.

C

C

Cy

Cx

By

Ay

Bx Ax

Cy

Cx

A B

C

Ax Bx

Ay

By

Given A, A

and B, B

Find components*

Ax = A cos A, Ay = A sin A

Bx = B cos B, By = B sin B

Add like components

Cx = Ax + Bx ,

Cy = Ay + By

Return to magnitude and

direction format

22

yx CCC

x

y

CC

C1tan **

3-7

Now let’s use the concept of vectors to extend the kinematical variables to more dimensions.

Average velocity is the displacement over the time,

tav

rv

Instantaneous velocity is

tt

rv

0lim

The velocity is tangent to the path of the particle.

The average acceleration is

tav

va

Instantaneous acceleration is

tt

va

0lim

“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 y-

components. The basic rule is

WE DO NOT DEAL WITH VECTORS. WE DEAL WITH THEIR COMPONENTS.

3-8

For the velocity, we have

t

yv

t

xv avyavx

,,

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,

tavvv xixfxx

becomes two equations:

tavvvv yiyfyyx and0 .

tvvx ixfx )(21

becomes two equations as well:

tvvytvx iyfyx )(and21 .

2

21 )( tatvx xix

becomes

2

21 )(and tatvytvx yiyx .

xavv xixfx 222

becomes

yavvvv yiyfyixfx 2and0 2222 .

Summary (see page 67):

x-axis : ax = 0 y-axis: constant ay Equation

0 xv tavvv yiyfyy (3-19)

tvx x tvvy iyfy )(21 (3-20)

2

21 )( tatvy yiy (3-21)

yavv yiyfy 222 (3-22)

3-9

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.

3-10

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.

Solution:

1. When the ball hits the ground, y = 0.

)(0

)(

21

2

21

tavt

tatvy

yiy

yiy

Since the only way a product can equal zero is when one of the factors equals zero,

0)(or0 21 tavt yiy

0

5

10

15

20

25

30

35

40

45

0 20 40 60 80 100 120

iyv

ixv

iv

= 55o

3-11

The first condition tells us that the golf ball starts from the ground. The second gives us

the time of flight, tf

y

iy

f

iyy

yiy

a

vt

vta

tav

2

0)(

21

21

This is more useful if we use ay = −g and viy = vi sin (see the diagram below the

trajectory plot).

s85.5

m/s8.9

55sin)m/s35(2

sin2

2

2

g

v

a

vt

i

y

iy

f

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.

g

vt

tgv

tavv

ih

hi

yiyfy

sin

sin0

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

2

21

2

21

2

21

sinsinsin

)(sin

)(

g

vg

g

vv

tgtvh

tatvy

iii

hhi

yiy

3-12

m/s9.41

)m/s8.9(2

55sin)m/s35(

2

sin

sin

2

1sin

2

22

22

2

2222

g

v

g

vg

g

vh

i

ii

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).

g

v

g

vv

tvR

tvx

i

ii

fi

ix

cossin2

sin2cos

cos

2

This can be rewritten using the identity sin 2 = 2sin cos ,

m117

m/s8.9

)]55(2sin[)m/s35(

2sin

2

2

2

g

vR i

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 ,

2cos)290sin()]45(2sin[

and

2cos)2cos()290sin()]45(2sin[

3-13

The range for 1 = 45º + is the same for 2 = 45º − . Another way is to say if

90)45()45(21

the ranges are the same!

4. Where is the ball at 2 seconds? Its horizontal position is found using

m2.40

s255cos)m/s35(

cos

fi

ix

tv

tvx

Its vertical position is

m7.37

)s2)(m/s8.9()s2(55sin)m/s35(

)(sin

)(

22

21

2

21

2

21

tgtv

tatvy

i

yiy

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,

s85.3s00.2s85.5 t

3-14

Another way is to use the symmetry of the y-component of the velocity. The y-

components of the velocities at the same heights have the same magnitudes but opposite

signs. At 2 s

m/s07.9

)s2)(m/s8.9(55sin)m/s35(

sin

2

tgvv

tavv

ify

yiyfy

When is the speed −9.07 m/s?

s85.3

m/s8.9

55sinm/s)35(m/s07.9

sin

sin

2

g

vvt

tgvv

tavv

ify

ify

yiyfy

Finally, the worst way is to just solve for t using the quadratic formula,

s86.3,s00.2

8.9

11.967.28

)9.4(2

)7.37)(9.4(4)67.28()67.28(

2

4

0m7.37)m/s67.28())(m/s9.4(

0m7.3755sin)m/s35())(m/s8.9(

))(m/s8.9(55sin)m/s35(m7.37

)(sin

)(

2

2

22

22

21

22

21

2

21

2

21

a

acbbt

tt

tt

tt

tgtv

tatvy

i

yiy

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.)

3-15

5. To find when the ball reaches 20 m we use the quadratic equation again,

s04.5,s81.0

8.9

74.2067.28

)9.4(2

)20)(9.4(4)67.28()67.28(

2

4

0m20)m/s67.28())(m/s9.4(

0m2055sin)m/s35())(m/s8.9(

))(m/s8.9(55sin)m/s35(m20

)(sin

)(

2

2

22

22

21

22

21

2

21

2

21

a

acbbt

tt

tt

tt

tgtv

tatvy

i

yiy

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 g

vt i

f

sin2

Time to height fi

h tg

vt 2

1sin

Maximum height g

vh i

2

sin22

Range g

vR i 2sin

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.

3-16

The time it takes to hit the ground is found from the equation for motion in the y-direction.

s47.2

s/m8.9

m302

2

)(0

)(

2

2

21

2

21

g

yt

tg

tatvy yiy

The ball is 50 m from the base.

m/s2.20

s47.2

m50

t

xv

tvx

x

x

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.

y

x

y

x

vi

3-17

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.

TGWTWG vvv

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:

m/s11

m/s)10(m/s)1(

TGxWTxWGx vvv

Greg sees Wanda walking to the right at 11 m/s. What happens when she walks back to her seat?

m/s9

m/s)10(m/s)1(

TGxWTxWGx vvv

According to Greg, Wanda is walking at 9 m/s to the right.

Hopefully, this is pretty easy. But what about this?

3-18

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.

WSRWRS vvv

The rowboat is to head directly to the west.

Take components.

WSyRWyRSyWSxRWxRSx vvvvvv and

The diagram is the key to solving relative velocity problems. For the x-component,

cos

0cos

RW

RWRS

WSxRWxRSx

v

vv

vvv

The y-component,

sin

sin0

RWWS

WSRW

WSyRWyRSy

vv

vv

vvv

Our unknowns are and vRS. From the y-component equation,

3-19

8.41

667.0m/s9.0

m/s6.0sin

sin

RW

WS

RWWS

v

v

vv

From the x-component equation.

m/s67.0

8.41cosm/s)90.0(

cos

RWRS vv

The boat must point 41.8º N of W upstream. Its speed across the water is 0.67 m/s.