1. Kinematics Kinematic Equations

2. Objectives
Determine the values, with correct signs, for the kinematic variables needed to quantitatively analyze horizontal motion.
Perform necessary unit conversions
Solve problems for horizontal motion using the kinematic equations
Solve problems using the quadratic equation as needed

3. Assumptions Assume the simplest environment
Problem statements often simplified and shortened by leaving out information regarding common assumptions.
Example: Every problem does not need to state that the motion occurs on Earth.
What are some other assumptions that are left out of problem statements.
Assume the simplest environment
Assume horizontal motion, unless inclines or vertical motion are mentioned.
Assume no friction, unless the problem mentions friction or implies its presence.
Assume no air resistance, unless mentioned in the problem.
If starting location and direction are not given, then you get to set them. Select an initial position and direction of motion that is mathematically easy.
Start objects at the origin, x0 = 0 . Substituting into Δx = x − x0 results in Δx = x .
Then how far the object moves, Δx , will have the same value as final position, x .
Set initial motion to the right (the +x direction) resulting in more + variables.

4. Acceleration Equation
In a previous lesson acceleration was defined as the rate of change in velocity, and the following equation was introduced.
This equation can be expanded as follows
In most cases initial time is zero
Which rearranges into one of the key kinematic equations
Note: This can be rewritten as
Which matches with
For a velocity-graph: Slope = acceleration
y-intercept = initial velocity

5. Additional Kinematic Equations
In the Galileo’s Incline lab an equation relating displacement and acceleration was introduced. However, this equation solves only for a special case where an object starts at the origin and accelerates from rest.
A more complete version of this equation allows for an initial position other than the origin, and for objects that have an initial velocity. This version is our next kinematic equation.
The first two kinematic equations can be combined to create a third kinematic equation.
There are also two ways to solve for average velocity in acceleration problems.

6. Clarification: Vectors and the Kinematic Equations
The quantities in the kinematic equations are vectors, yet AP Central’s equations have these quantities written in italics (scalar values) rather than as vectors. WHY?
Technically position is a vector measured from the origin to the location of the object, and in three dimensional space it can be annotated either as s or r .
This means that displacement in three dimensional space is either  s or  r .
If the equations are written in vector form they might appear as follows.
However, vectors can point in any direction in three dimensional space.
Adding vectors that point in radically different directions can be very complicated. We saw that when vectors were perpendicular, vector addition required Pythagorean theorem. Vector addition at angle other than 90o is more complicated.
How do we simplify this?

7. Clarification: Vectors and the Kinematic Equations
Convert the three dimensional vectors
into scalar quantities, which allow for both + and  signs and can use normal addition.
For this to occur the vectors must be parallel. This is why AP Central indicates these quantities with x subscripts. They are indicating that the vectors must all lie along the same axis, in this case the x-axis

8. Clarification: Vectors and the Kinematic Equations
Are we limited to only the x-axis?
No. Simply change the subscripts
Example: For vertical motion change the subscripts to y .
How do we handle diagonal motion, such as up and down inclines?
You decide where the coordinate axis is positions and how it is oriented.
Tilt the axis
Do we need to include the x and y subscripts when writing equations?
Yes and No.
In problems involving both horizontal and vertical motion the subscripts are an absolute necessity.
However, if problems move in only one dimension, then they are not necessary.
I class, in these PowerPoint presentations, and all my documents I will usually only include the axis subscripts when they are vital to the solution. +x +x

9. Clarification: Vectors and the Kinematic Equations
The previous slides explained that parallel kinematic vector quantities were changed into scalars quantities for vector addition purposes.
What does these equations have to do with vector addition?
The kinematic equations are adding like vectors.
To add vectors using ordinary addition, they must be of the same type and they must all be parallel to one another.

10. Kinematic Equations When Used No final speed No time No position
Complete version
Rewritten in terms displacement (x = x  x0)
Object starts at rest
(v0 = 0)
Object has constant velocity
(a = 0)

11. Displacement versus Initial and Final Position
The more complete form of the following equations is printed on the equation sheet provided by AP Central for the AP Exam.
However, most problems will specify how far the object moves, x , rather than its starting and ending locations , x0 and x . As a result, I prefer the modified version of these equations that addresses displacement. You are more likely to see me using the following versions of these equations.
Here is something else to consider. In most problems the object is initially located at the origin. In these cases x0 = 0 , and then these equations simplifies to
They are all acceptable. The original complete versions, at the top, only needs to be used if the object’s starting location is specifically stated as not being at the origin.

12. Kinematic Variables Symbol Variable Positive Negative Zero x0
Initial position
Object is physically located right of the origin
Object is physically located left of the origin
Object is physically located at the origin x
Final position x
Displacement
Object is moved right
Object is moved left
Object is stationary v0
Initial velocity
Initial motion is to the right
Initial motion is to the left
Initially at rest v
Final velocity
Final motion is to the right
Final motion is to the left
Comes to a stop a
Acceleration
Usually if speeding up
Usually if slowing down
Constant velocity t
Elapsed time
Always in this class
N/A to this course

13. Making a Variable List Here is your dilemma
Problems will be word problems.
They will seem vague, as they expect you to make certain assumptions
They can mention up to 6 variables that can be positive, negative, or zero
They may mention extraneous variables to distract you
Sometimes numbers are given, and sometimes values are implied in words
Values will not always be in correct SI units, and may require metric conversion.
How do you keep it all straight?
Try a making a variable list as you read the problem.
A variable list turns a word problem into a math problem.
Include zero quantities and plus and minus signs.
Perform the metric conversions in the list and before you start calculating.
The variable list makes it easier to select the correct equation.
In addition, it can helps identify if you overlooked a variable and need to reread the problem. This is particularly true for variables implied in writing.
The following examples start with, and demonstrate, how to use variable list.

14. An object initially at rest accelerates uniformly at 6
An object initially at rest accelerates uniformly at 6.0 m/s2 while moving 250 m. Determine the time for this motion.
Example 1
Look for hidden zero quantities, and picture a coordinate axis to assist in determining the correct sign of each variable.
Δx = 250 m Given in the text of the problem
v0 = m/s Hidden in the words, “initially at rest”
v = Never given
a = 6.0 m/s2 Given in the text of the problem
t = ? The unknown quantity you are solving for

15. An object initially at rest accelerates uniformly at 6
An object initially at rest accelerates uniformly at 6.0 m/s2 while moving 250 m. Determine the time for this motion.
Example 1
Deciding on an equation
1. What type of motion is involved?
If you answer constant velocity, use
If you answer acceleration continue.
2. Is time mentioned in any way?
In not, then try
If time is mentioned continue.
3. Is displacement or final velocity mentioned?
If displacement is mentioned, try
If final velocity is mention, try
Δx = 250 m
v0 = m/s
v =
a = 6.0 m/s2
t = ?
Accelerating: Time and displacement are mentioned

16. Example 1 Solve showing the following work
An object initially at rest accelerates uniformly at 6.0 m/s2 while moving 250 m. Determine the time for this motion.
Example 1
Solve showing the following work
1. Write the original equation (or an equivalent)
2. Rearrange to solve for requested variable
3. Show substitution of given values
4. Solve, box answer, and include correct units
Δx = 250 m
v0 = m/s
v =
a = 6.0 m/s2
t = ?
Accelerating: Time and displacement are mentioned

17. An object moving at 32.0 m/s decelerates to a stop in a distance of 240 m. Determine the objects acceleration.
Example 2
Acceleration has two signs.
It is a vector, so direction of motion influences its sign. Since no direction was given, I assumed the object was moving in the +x direction. It never reversed direction, so direction remained positive.
Acceleration is also a rate. If velocity is increasing (speeding up) acceleration is positive, and if decreasing (slowing) it is negative.
Here the object was decelerating in the positive direction. − + a = −a .
Δx = 240 m
v0 = 32.0 m/s
v = m/s
a =
t =
Accelerating, and time is not mentioned.

18. An object with an initial velocity of 10 m/s accelerates to a speed of 20 m/s in a distance of 100 m. Determine the elapsed time during this motion
Example 3
Find acceleration
1. Original Equation
2. Rearrange
3. Substitute
4. Solve
Find time
1. Original Equation
2. Rearrange
3. Substitute
4. Solve
Δx = 100 m
v0 = 10 m/s
v = 20 m/s
a =
t = ?
Find acceleration
Then find time

19. An object uniformly accelerates from rest reaching a final speed v while moving a distance x . Determine its speed when it has moved a distance 2x . Answer in terms of v .
Example 4
Δx =
v0 =
v =
a =
t =
No values, but you can still use a variable list. Simply check of any variable mentioned or implied.
An object uniformly accelerates check off a
from rest v0 = 0
reaching a speed v check off v
while moving a distance x check of Δx
0 (rest)
Accelerating. Time not mentioned.
Substitute v0 = 0 , and rearrange in terms of v .
How would final speed change if x is doubled to 2x ?
Double x in the derived equation and determine the coefficient needed in front of v to maintain the equality.
Doubling x increases the right side by root two. The left side must also be multiplied by root two.

20. Example 5 First well try solving this using the quadratic equation.
An object is moving in the +x direction at 25 m/s. At the instant it passes through the origin it encounters an acceleration of m/s2 directed opposite its motion. The acceleration ceases to act when the object is 6.0 m to the left of the origin. Determine the elapsed time while the acceleration acts.
Δx = −6.0 m
v0 = +25 m/s
v =
a = −3.0 m/s2
t = ?
Begin by identifying key variables and their correct signs.
First well try solving this using the quadratic equation.
Negative time makes no sense in real life.
Ignore the negative, and choose the positive value.

21. Example 5 Now let’s try an alternate method.
An object is moving in the +x direction at 25 m/s. At the instant it passes through the origin it encounters an acceleration of m/s2 directed opposite its motion. The acceleration ceases to act when the object is 6.0 m to the left of the origin. Determine the elapsed time while the acceleration acts.
Δx = −6.0 m
v0 = +25 m/s
v =
a = −3.0 m/s2
t = ?
Now let’s try an alternate method.
Problems can often be solved using more than one method.
You should be able to solve this problem using the quadratic equation.
However, the alternate method does have one advantage. Final velocity is calculated, as well as the elapsed time.
If the object is to the left of the origin it must be moving left. Choose the negative velocity.

22. Multiphase Kinematics Problems
Some kinematics problems are actually several different problems in one.
How are these identified?
If a problem alludes to more than one acceleration, more than one time interval, and/or more than one displacement it is a multiphase problem.
There are several ways to do this, but beginning students may find it easier to solve each part separately.
If asked for the total displacement: Try solving for the displacement in each phase separately, and then adding them together.
If asked for the total time: Try solving for the time in each phase separately, and then adding them together.

23. Example 6 I II III  x 1200 m 225 m v0 0 m/s v a 3.0 m/s2 t 10 s
A car is at rest at a traffic signal. When the light turns green the car accelerates at 3.0 m/s2 for 10 s. Then the driver maintains a constant velocity for 1.2 km. Finally the car slows to a stop in a distance of 225 m. Determine total displacement and elapsed time. I II
III
 x
1200 m
225 m v0
0 m/s v a
3.0 m/s2 t
10 s
Phase I
Accelerates at 3.0 m/s2
Phase II
Constant velocity, a = 0 m/s2
Phase III
Slows to stop. An unspecified acceleration that stops the car, v = 0 m/s
If a problem alludes to more than one acceleration, more than one time interval, and/or more than one displacement it is probably a multiphase problem.
This type of variable list is extremely useful.

24. Example 6 I II III  x 1200 m 225 m v0 0 m/s v a 3.0 m/s2 t 10 s Total
A car is at rest at a traffic signal. When the light turns green the car accelerates at 3.0 m/s2 for 10 s. Then the driver maintains a constant velocity for 1.2 km. Finally the car slows to a stop in a distance of 225 m. Determine total displacement and elapsed time. I II
III
 x
1200 m
225 m v0
0 m/s v a
3.0 m/s2 t
10 s
Total
In this problem we are solving for
Total displacement
xtotal = x1 + x2 + x3
Total elapsed time
ttotal = t1 + t2 + t3
Add another column

25. Important: Example 6 I II III  x 1200 m 225 m v0 0 m/s v a 3.0 m/s2 t
A car is at rest at a traffic signal. When the light turns green the car accelerates at 3.0 m/s2 for 10 s. Then the driver maintains a constant velocity for 1.2 km. Finally the car slows to a stop in a distance of 225 m. Determine total displacement and elapsed time. I II
III
 x
1200 m
225 m v0
0 m/s v a
3.0 m/s2 t
10 s
Total
Important:
The key to all multiphase problems throughout the course:
The final values of phase I become the initial values for phase II, etc.
The key value to transfer to the next phase will vary depending on the type of problem you are solving. However, in multiphase kinematics problems the key values are the velocities. The final velocity of phase I becomes the initial velocity of phase II, etc.

26. Example 6 I II III  x 1200 m 225 m v0 0 m/s v a 3.0 m/s2 t 10 s Total
A car is at rest at a traffic signal. When the light turns green the car accelerates at 3.0 m/s2 for 10 s. Then the driver maintains a constant velocity for 1.2 km. Finally the car slows to a stop in a distance of 225 m. Determine total displacement and elapsed time. I II
III
 x
1200 m
225 m v0
0 m/s v a
3.0 m/s2 t
10 s
Total
Start by finding the missing values in Phase I
Then add these values to the variable list
150 m
30 m/s

27. Example 6 I II III  x 1200 m 225 m v0 0 m/s v a 3.0 m/s2 t 10 s Total
A car is at rest at a traffic signal. When the light turns green the car accelerates at 3.0 m/s2 for 10 s. Then the driver maintains a constant velocity for 1.2 km. Finally the car slows to a stop in a distance of 225 m. Determine total displacement and elapsed time. I II
III
 x
1200 m
225 m v0
0 m/s v a
3.0 m/s2 t
10 s
Total
The final velocity in phase I becomes the initial velocity in phase II.
Phase II is constant velocity. As a result the final and initial velocities will be equal in this phase.
The final velocity in phase II becomes the initial velocity in phase III.
150 m
30 m/s
30 m/s
30 m/s
30 m/s

28. Example 6 t = 40 s I II III  x 1200 m 225 m v0 0 m/s v a 3.0 m/s2 t
A car is at rest at a traffic signal. When the light turns green the car accelerates at 3.0 m/s2 for 10 s. Then the driver maintains a constant velocity for 1.2 km. Finally the car slows to a stop in a distance of 225 m. Determine total displacement and elapsed time. I II
III
 x
1200 m
225 m v0
0 m/s v a
3.0 m/s2 t
10 s
Total
Solve for the last missing quantity in phase II
Add this to the list
150 m
30 m/s
30 m/s
30 m/s
30 m/s
40 s
t = 40 s

29. Example 6 t = 15 s I II III  x 1200 m 225 m v0 0 m/s v a 3.0 m/s2 t
A car is at rest at a traffic signal. When the light turns green the car accelerates at 3.0 m/s2 for 10 s. Then the driver maintains a constant velocity for 1.2 km. Finally the car slows to a stop in a distance of 225 m. Determine total displacement and elapsed time. I II
III
 x
1200 m
225 m v0
0 m/s v a
3.0 m/s2 t
10 s
Total
Solve for the missing quantities in phase III.
Add them to the list.
150 m
30 m/s
30 m/s
30 m/s
30 m/s
−2.0 m/s2
40 s
15 s
t = 15 s

30. Example 6 I II III  x 1200 m 225 m v0 0 m/s v a 3.0 m/s2 t 10 s Total
A car is at rest at a traffic signal. When the light turns green the car accelerates at 3.0 m/s2 for 10 s. Then the driver maintains a constant velocity for 1.2 km. Finally the car slows to a stop in a distance of 225 m. Determine total displacement and elapsed time. I II
III
 x
1200 m
225 m v0
0 m/s v a
3.0 m/s2 t
10 s
Total
Determining total displacement and total elapsed time should be obvious
xtotal = x1 + x2 + x3
xtotal = 1575 m
ttotal = t1 + t2 + t3
ttotal = 65 s
150 m
1575 m
30 m/s
30 m/s
30 m/s
30 m/s
−2.0 m/s2
40 s
15 s
65 s
This table version of the variable list is especially useful if one or more of the total values is given, and an initial quantity is needed.
The table clearly shows what is missing and dictates the order of the equations for you.