1. College Physics for the AP® Physics 1 Course
Tom Pfeiffer/VolcanoDiscovery/Getty Images
Vectors
Centripetal Motion
Motion in Multiple Dimensions
Projectile Motion
College Physics for the AP® Physics 1 Course
Presentation Created by Martin Kirby

2. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
1-D, 2-D, and 3-D Motion
x-axis
One-dimensional motion
Linear motion, studied in previous chapter.
1-D
x-axis
y-axis
Two-dimensional motion
2-D
x-axis
y-axis
z-axis
Three-dimensional motion
3-D
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

3. Vectors and Scalar Quantities 1
Any quantity that has only magnitude (and no direction) is called a
scalar quantity.
Examples:
Distance Speed
More examples
Mass
Temperature
Money
Energy
Power IQ
Hat Size
Volume
Area
Number of Senators
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

4. Vectors and Scalar Quantities 2
Any quantity that has magnitude and direction is called a vector quantity.
Examples of vector quantities:
Force… try to create a single force that has no direction!
Linear Displacement … without direction it’s distance.
Linear Velocity…without direction it’s speed.
Linear Acceleration... without direction it’s just speeding
up or slowing down.
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

5. 3 carrots – 1 carrot = 2 carrots 4 kg + 2 kg = 6 kg 6˚C – 10˚C = – 4˚C
Scalar Math
You have been doing scalar math since you were very young. Your scalar math calculations have always used quantities that have magnitude, and no direction.
Examples:
4 + 2 = 6
8 – 3 = 5
2 × 7 = 14
12 ÷ 6 = 2
= 20
$5 + $2 = $7
3 carrots – 1 carrot = 2 carrots
4 kg + 2 kg = 6 kg
6˚C – 10˚C = – 4˚C
… now we will learn a little vector math.
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

6. How to Represent a Vector
A straight arrow is used to represent a vector quantity.
The length of the arrow represents the magnitude (size) of the vector quantity.
The direction in which the arrow is pointing represents the direction of the vector quantity.
length = magnitude
direction = direction
AN ARROW REPRESENTS A VECTOR QUANTITY
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

7. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Examples of Vectors
= 2 m/s to the east
= 3 m/s to the east
= 4 m/s to the west
= –4 m/s to the east
= 2 m/s2 to the east N S E W
= 1 m to the west
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

8. Question: Replacing One Vector with Two
= 5 m/s
𝑣 NORTH
30˚ N S E W
𝑣 EAST
A car is moving along a level road at a velocity of 5 m/s, 30˚ north of east
How fast is the car moving north?
How fast is the car moving east?
Answers on next slide…
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

9. Answer: Replacing One Vector with Two
This trig. information is provided on the AP Physics 1 equation page.
= 5.00 m/s
𝑣 𝐧𝐨𝐫𝐭𝐡
30˚
𝑣 𝐞𝐚𝐬𝐭
The two replacement vectors are called components.
cos 30˚ = 𝑣 𝐞𝐚𝐬𝐭 /5.00
𝑣 𝐞𝐚𝐬𝐭 = (5.00) cos 30˚
𝑣 𝐞𝐚𝐬𝐭 = (5.00)(0.866)
𝑣 𝐞𝐚𝐬𝐭 = 4.33 m/s
To find 𝑣 𝐞𝐚𝐬𝐭
sin 30˚ = 𝑣 𝐧𝐨𝐫𝐭𝐡 / 5.00
𝑣 𝐧𝐨𝐫𝐭𝐡 = (5.00) sin 30˚
𝑣 𝐧𝐨𝐫𝐭𝐡 = (5.00)(0.5)
𝑣 𝐧𝐨𝐫𝐭𝐡 = 2.50 m/s
To find 𝑣 𝐧𝐨𝐫𝐭𝐡
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

10. Find the result of adding these two vectors….
Adding Two Vectors 1 N S E W
Find the result of adding these two vectors….
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

11. Adding Two Vectors 2
To add the two vectors shown in the previous slide, put them together so that they are head to tail.
resultant
Then complete a triangle by drawing a vector that starts at the free tail and ends at the free arrowhead. This is the answer!… the result of adding two vectors.
The result of adding vectors is called the resultant.
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

12. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Adding Two Vectors 3 𝐵 𝐴
The blue resultant vector replaces the two red vectors that have been added together.
resultant 𝐶
𝑨 + 𝑩 = 𝑪
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

13. Subtracting Two Vectors
resultant 𝐶
To subtract 2 vectors: 𝐵 𝐵
𝑨 − 𝑩 = 𝑪 𝐵 𝐵 𝐵 𝐵 𝐵 𝐵 𝐵
- 𝐵 𝐵 𝐵 𝐵 𝐵 𝐴
Reverse direction of 𝐵 ,
then add the two vectors:
𝑨 +(− 𝑩 )= 𝑪
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

14. Compare Vector Addition and Subtraction
𝑨 − 𝑩 = 𝑪 𝐴 𝐵
resultant 𝐶
𝑨 +(− 𝑩 )= 𝑪 𝐴
resultant 𝐶
− 𝐵
𝑨 + 𝑩 = 𝑪
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

15. Example of Using a Vector Diagram 1
RIVER VELOCITY:
3 m/s to the West
This slide shows a river moving with a velocity of 3 m/s to the West.
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

16. Example of Using a Vector Diagram 2
River is not moving
This slide shows a person (red dot) swimming straight across a stationary river with a velocity of 4 m/s to the North.
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

17. Example of Using a Vector Diagram 3
SWIMMER VELOCITY:
4 m/s to the North
RIVER VELOCITY:
3 m/s to the West
This slide shows a person swimming
(with a velocity of 4 m/s to the North)
straight across a river
(that has a velocity of 3 m/s to the West).
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

18. Example of Using a Vector Diagram 4
SWIMMER VELOCITY:
4 m/s to the North
RIVER VELOCITY:
3 m/s to the West
4 m/s
3 m/s
Resultant velocity
4 m/s
3 m/s
5 m/s
Resultant velocity
4 m/s
3 m/s
Pythagorean theorem:
Resultant2 = = 25
Resultant = 𝟐𝟓 = 5 m/s
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

19. Example of Using a Vector Diagram 5
SWIMMER VELOCITY:
4 m/s to the North
RIVER VELOCITY:
3 m/s to the West
4 m/s
3 m/s
Resultant velocity
4 m/s
3 m/s
5 m/s
Resultant velocity
4 m/s
3 m/s
Pythagorean theorem:
Resultant2 = = 25
Resultant = 𝟐𝟓 = 5 m/s
tan θ = 4/3
tan θ = 1.33
θ = 53.1˚ N of W θ
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

20. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Projectile Motion 1
Projectile motion: the motion of an object that is launched with a velocity that has both horizontal and vertical components and then falls freely.
We are going to study projectile motion as it provides us with opportunities to work with motion in two dimensions.
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

21. constant acceleration
Projectile Motion 2 v
constant velocity
When any object is thrown near Earth’s surface, its motion can be analyzed as two separate motions, whose only connection is time.
When an object is thrown:
The object has constant horizontal velocity.
The object has constant vertical acceleration, g.
constant acceleration g
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

22. Ball thrown on planet with no gravity
v= +50 m/s
v= +50 m/s
v= +50 m/s
v= +50 m/s
v= +50 m/s
v= +50 m/s
v= +50 m/s
v= +50 m/s
v= +50 m/s
0 s
1 s
2 s
3 s
4 s
5 s
6 s
7 s
∆x=+50m
∆x=+50m
∆x=+50m
∆x=+50m
∆x=+50m
∆x=+50m
∆x=+50m
A ball is thrown at 50 m/s on a planet with no gravity.
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

23. Ball released from rest on planet that has gravity
constant
vertical acceleration
g = –10 m/s2
0 s
v = 0 m/s
∆y = –5 m
1 s
v = –10 m/s
∆y = –15 m
2 s
v = –20 m/s
∆y = –25 m
v = –30 m/s
3 s
∆y = –35 m
v = –40 m/s
4 s
∆y = –45 m
A ball is released from rest (0 m/s) on a planet that has gravity.
5 s
v = –50 m/s
∆y = –55 m
6 s
v = –60 m/s
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

24. The Two Previous Motions Shown Together
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 0 m/s
v = –10 m/s
v = – 20 m/s
v = – 30 m/s
v = – 40 m/s
v = – 50 m/s
v = – 60 m/s
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

25. The previous motions combined
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 50 m/s
v = 0 m/s
v = -10 m/s
v = -20 m/s
v = -30 m/s
v = -40 m/s
v = -50 m/s
v =- 60 m/s
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

26. Combined Motions with Vector Diagrams
v = 50 m/s
v = 50 m/s
v = – 10 m/s
Constant horizontal velocity
v = 50 m/s
∆y = –5 m
∆y = –15 m
v = 50 m/s
v = – 20 m/s
∆y = –25 m
v = 50 m/s
v = – 30 m/s
∆y = –35 m
v = 50 m/s
v = –40 m/s
∆y = –45 m
Constant vertical acceleration
a = -g = –10 m/s2
v = 50 m/s
∆y = –55 m
v = – 50 m/s
∆x=+50m
∆x=+50m
∆x=+50m
∆x=+50m
∆x=+50m
∆x=+50m
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

27. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Projectile Question
A tennis ball is thrown horizontally off a cliff with a speed of 30 m/s. Assume that g = 10 m/s2. Calculate the following:
The ball’s horizontal velocity at t = 4 s
The ball’s vertical velocity at t = 4 s
The ball’s speed at t = 4 s
The angle at which the ball is moving, relative to the horizontal at t = 4 s
The horizontal displacement of the ball at t = 4 s
The vertical displacement of the ball at t = 4 s
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

28. Projectile Question: Answers to (a) and (b)
A tennis ball is thrown horizontally off a cliff with a speed of 30 m/s. Assume that g = 10 m/s2. Calculate the following:
The ball’s horizontal velocity at t = 4.0 s
The ball has constant horizontal velocity, therefore it is always moving at 30 m/s horizontally to the east, until it hits the ground.
b) The ball’s vertical velocity at t = 4.0 s
vo = 0
v = ?
∆y =
a = –10 m/s2
∆t = 4.0 s
a = (v – vo) / ∆t
–10 m/s2 = (v – 0 ) / (4.0 s)
v = –40 m/s
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

29. Projectile Question: Answers to (c) and (d)
A tennis ball is thrown horizontally off a cliff with a speed of 30 m/s.
Assume that g = 10 m/s2. Calculate the following:
The ball’s horizontal velocity at t = 4.0 s 30 m/s
The ball’s vertical velocity at t = 4.0 s –40 m/s
The ball’s speed at t = 4.0 s
v2 = (30 m/s)2 + (–40 m/s)2
v2 =
v = = m/s
30 m/s θ v
d) The angle at which the ball is moving at t = 4 s, relative to the horizontal
–40 m/s
tan θ = (–40 m/s) / (30 m/s)
tan θ = –1.33
θ = 53.1˚
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

30. Projectile Question: Answers to (e) and (f)
A tennis ball is thrown horizontally off a cliff with a speed of 30 m/s.
Assume that g = 10 m/s2. Calculate the following:
horizontal velocity of ball at t = 4.0 s 30 m/s
vertical velocity of ball at t = 4.0 s -40 m/s
speed of ball at t = 4.0 s 50 m/s
The angle at which the ball is moving at t = 4 s, relative to the horizontal 53.1˚
e) The horizontal displacement of the ball
at t = 4.0 s
f) The vertical displacement of the ball
at t = 4.0 s
vo = 30 m/s
v = 30 m/s
∆xhoriz = ?
ahoriz = 0 m/s2
∆t = 4.0 s
vo = 0
v = –40 m/s
∆yvert = ?
avert = –10 m/s2
∆t = 4.0 s
same time
∆xhoriz = vot + ½a∆t2
∆xhoriz = (30)(4 s) + ½(0 m/s2)(4.0 s)2
∆xhoriz = m = 120 m
∆yvert = vot + ½a∆t2
∆yvert = (0)(4 s) + ½(–10 m/s2)(4.0 s)2
∆yvert = 0 + (-80 m)= -80 m
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

31. Projectile Question: Animated
v = 30 m/s
v = 30 m/s
v = – 10 m/s
Constant horizontal velocity
v = 30 m/s
∆y = –5 m
∆y = – 15 m
v = 30 m/s
v = – 20 m/s
∆y = –25 m
v = 30 m/s
v = – 30 m/s
∆y = –35 m
v = 30 m/s
v = –40 m/s
θ = 53.1˚
Constant vertical acceleration
a = -g = –10 m/s2
v = 50 m/s
∆x=+30 m
∆x=+30 m
∆x=+30 m
∆x=+30 m
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

32. Chapter 3 Summary 1 𝐴 𝐵 𝐴 𝐶 − 𝐵 Addition 𝐶 Subtraction
Scalar quantity: has only magnitude (size)
Vector quantity: has magnitude and direction
Scalar math: ordinary math
Vector math: a vector quantity is represented using an arrow, whose length represents magnitude, and whose arrowhead represents direction. Vector math uses these arrows. 𝐴
resultant 𝐶
− 𝐵 𝐴 𝐵
resultant 𝐶
𝑨 − 𝑩 = 𝑪
𝑨 + 𝑩 = 𝑪
𝑨 +(− 𝑩 )= 𝑪
Addition
Subtraction
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE

33. Constant vertical acceleration Constant horizontal velocity
Chapter 3 Summary 2
Constant vertical acceleration
Constant horizontal velocity
Projectile motion:
Objects that are in projectile motion have constant horizontal velocity and constant vertical acceleration.
The only link between the horizontal and vertical motions is time.
COLLEGE PHYSICS FOR THE AP®  PHYSICS 1 COURSE