Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 1

Last time we looked at position vs. time and acceleration vs. time graphs. Since the

instantaneous velocity is

t

xv

tx

0lim

the (instantaneous) velocity can be found from the slope of the position vs. time graph. In a

similar way, since

t

va x

tx

0lim

the acceleration can be found from the slope of the velocity vs. time graph. (By acceleration we

refer to instantaneous acceleration.)

Assuming that the acceleration is constant, we can find the four (very) useful relations.

tavvv xixfxx

This is a rearrangement of the definition of ax. No displacement in the equation!

tvvx ixfx )(21

The displacement equals the average velocity times the elapsed time. No acceleration in the

equation!

2

21 )( tatvx xix

The displacement equals the sum of the displacement due to the initial velocity and the

displacement due to the acceleration. No final speed!

xavv xixfx 222

The difference in the squares of the velocities equals twice the product of the acceleration and

the displacement. No time!

A very common situation with constant acceleration is free fall. If we neglect air resistance, all

objects fall with the same acceleration, regardless of the object’s mass or state of motion.

Surprised? Many people were surprised when Galileo discovered this fact.

A video demonstration: http://www.youtube.com/watch?v=E43-CfukEgs

Consider the following. What if heavier objects accelerated faster? (From

http://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment):

Galileo arrived at his hypothesis by a famous thought experiment outlined in his

book On Motion. Imagine two objects, one light and one heavier than the other

one, are connected to each other by a string. Drop this system of objects from the

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 2

top of a tower. If we assume heavier objects do indeed fall faster than lighter

ones (and conversely, lighter objects fall slower), the string will soon pull taut as

the lighter object retards the fall of the heavier object. But the system considered

as a whole is heavier than the heavy object alone, and therefore should fall faster.

This contradiction leads one to conclude the assumption is false.

The acceleration due to gravity is denoted by g and has the value

2m/s8.9g

The direction is downwards, towards the center of the earth. In fact, it is how we define down!

To create the equations of vertical motion, we use our original equations and replace x with y.

To create equations for free fall, replace the ay with –g.

original y-direction free-fall

tavvv xixfxx tavvv yiyfyy tgvvv iyfyy

tvvx ixfx )(21 tvvy iyfy )(

21 tvvy iyfy )(

21

2

21 )( tatvx xix 2

21 )( tatvy yiy

2

21 )( tgtvy iy

xavv xixfx 222 yavv yiyfy 222 ygvv iyfy 222

But in general, motion is not just along a line. How do we deal with that?

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).

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 3

B A

A+B

B

A 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?

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?

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.

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 4

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.

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

yC

xC

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 5

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 add to 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

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?

=40o

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 6

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 62) 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.

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

and y-components (page 62-63) 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.

=50o

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 7

22

yx AAA

Problem-Solving Strategy: Adding Vectors Using Components (page 63) 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.

We can use unit vectors to write vectors in a compact way. We define �̂� (read “x hat”) as the

unit vector in the x direction and similarly for �̂�. The components of �⃗⃗� can be written as

yAxA yyxxˆandˆ AA

and

yAxA yxyxˆˆ AAA

Using unit vectors, the sum of �⃗⃗� and �⃗⃗� can be written as

yBAxBA

yBxByAxA

yyxx

yxyx

ˆˆ

ˆˆˆˆ

BA

Notice that by factoring out the �̂� and �̂�, the x-components are added together and the y-

components are added together.

The unit vector notation can be very helpful but it will not be used in this course. I have

included it since it might be used in PHY2054.

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 8

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 63

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 **

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 9

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 can 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.” (page 68)

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.

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 10

For the velocity, we have

t

yv

t

xv avyavx

,,

with similar definitions for the other parameters. (See pages 66-68.)

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.” (page 69)

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 69):

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)

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 11

Projectiles are a good example of this type of motion. Here ay = −g.

This motion is simultaneously constant velocity in the x-direction with constant acceleration in

the y-direction. The motion in the x direction is independent of the motion in the y direction.

Lesson 3: Free fall, Vectors, Motion in a plane (sections 2.6-3.5)

Lesson 3, page 12

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?

Solution: Treat each component separately. Find the time it takes for the ball to hit the ground

and use that to find the initial velocity.

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

y

x

y

x

vi