1. Kinematics in One Dimension
Chapter 2
Kinematics in One Dimension
Giancoli, PHYSICS,6/E © Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey

2. Displacement, Velocity and Acceleration
Module 2
Displacement, Velocity and Acceleration
Giancoli, Sec 2-1, 2, 3, 4, 8
AP Physics C Lesson 1
The following is an excellent lecture on this material.
Giancoli, PHYSICS,6/E © Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey

3. Reference Frames
Any measurement of position, displacement, velocity or acceleration must be made with respect to a defined reference frame—this is first step in problem solution.
Possible reference frames:
Window with up = + or –
Un-stretched net with up = + or -
Stretched net with up = + or –
Ground = not sufficient information
Module 2-1

4. Coordinate Axis
We will use a set of coordinate axis where x is horizontal and y is vertical +y -y +x -x
x x2
Many problems will be motion in one dimension so we will plot x vs. time.
Module 2-2

5. Displacement Displacement: change in position+y -y +x -x
x x2
Displacement: change in position
Displacement is a vector, so it has magnitude and direction. In one dimension we use + or minus sign to indicate direction.
Module 2-3

6. Don’t Confuse Displacement and Distance
A person walks 70 m East and 30 m West.
Distance traveled =
100 m
or m
40 m East
Displacement =
Module 2-4

7. Negative Displacement
In the figure below the displacement is negative.
A negative displacement may indicate motion toward the West or something else depending on the situation and the coordinate system chosen.
Module 2-5

8. Average Speed and Velocity
Video Analysis: c:\videos\WIM\Amuseprk\13.avi
Set steps to 6
Average velocity:
Measure the two ends of the video and calculate the average velocity (13m/s) can use calculator on screen
Instantaneous velocity – after you define on later slide, mark points, show graph and compare
Acceleration – show after it has been defined
Average velocity is a vector, so it has magnitude and direction. In one dimension we use + or minus sign to indicate direction.
Module 2-6

9. Example 1. An airplane travels east 3100 km at a velocity of 790 km/h, and then encounters a tailwind that boosts its velocity to 990 km/h for the next 2800 km. What was the total time for the trip? (Assume three significant figures)
Module 2-7

10. What was the average velocity of the plane for this trip?
Example 1 (continued) An airplane travels east 3100 km at a velocity of 790 km/h, and then encounters a tailwind that boosts its velocity to 990 km/h for the next 2800 km. What was the total time for the trip? (Assume three significant figures)
What was the average velocity of the plane for this trip?
Note: A simple average of v1 and v2 gives 890 km/h and is not correct
Module 2-8

11. Instantaneous Velocity
instantaneous velocity is defined as the average velocity over an infinitesimally short time interval.
Module 2-9

12. Graphical Analysis of Linear Motion
Module 2-10

13. Acceleration
Average Acceleration: change in velocity divided by the time taken to make this change.
Module 2-11

14. Example 2. A car traveling at 15. 0 m/s slows down to 5. 0 m/s in 5
Example 2. A car traveling at 15.0 m/s slows down to 5.0 m/s in 5.0 seconds. Calculate the car’s acceleration.
Coordinate System: + is to the right
Module 2-12

15. Acceleration
Instantaneous Acceleration: same definition as before but over a very short  t.
Module 2-13

16. Motion with Constant Acceleration Giancoli, Sec 2-1, 2, 3, 4, 8
Module 3
Motion with Constant Acceleration
Giancoli, Sec 2-1, 2, 3, 4, 8
Giancoli, PHYSICS,6/E © Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey

17. Motion with Constant Acceleration
Consider the case of a car that accelerates from rest with a constant acceleration of 15 m / s2.
We can make a table
t (s)
v ( m/s)
a ( m/ s2 ) 15 1 2 30 3 45 4 60 5 75
Module 3-1

18. Derivations
In the next 4 slides we will combine several known equations under the assumption that the acceleration is constant.
This process is called a derivation.
In general you will need to know the initial assumptions, the resultant equations and how to apply them.
You do not need to memorize derivations
But, I could ask you to derive an equation for a specific problem. This is very similar to an ordinary problem without a numeric answer.
Module 3-2

19. Motion at Constant Acceleration - Derivation
Consider the special case acceleration equals a constant: a = constant
Use the subscript “0” to refer to the initial conditions
Thus t0 refers to the initial time and we will set t0 = 0.
At this time v0 is the initial velocity and x0 is the initial displacement.
At a later time t, v is the velocity and x is the displacement
In the equations t1t0 and t2  t
Module 3-3

20. Motion at Constant Acceleration - Derivation
The average velocity during this time is:
The acceleration is assumed to be constant
From this we can write
Module 3-4

21. Motion at Constant Acceleration - Derivation
Because the velocity increases at a uniform rate, the average velocity is the average of the initial and final velocities
From the definition of average velocity
And thus
Module 3-5

22. Motion at Constant Acceleration
The book derives one more equation by eliminating time
The 4 equations listed below only apply when a = constant
Module 3-6

23. Example 3. A world-class sprinter can burst out of the blocks to essentially top speed (of about m/s) in the first 15.0 m of the race. What is the average acceleration of this sprinter, and how long does it take her to reach that speed? (Note: we have to assume a=constant)
Module 3-7

24. Example 4. A truck going at a constant speed of 25 m/s passes a car at rest. The instant the truck passes the car, the car begins to accelerate at a constant 1.00 m / s2. How long does it take for the car to catch up with the truck.
When the car catches the truck:
How far has the car traveled when it catches the truck?
Module 3-8

25. Quiz Calculate the average acceleration from t = 0 s to t = 30s.
A m/s2
B m/s2
C 15 m/s2
D none of the above
Ch 2

26. Quiz Calculate the acceleration from t = 0 s to t = 15 s. A 0 m/s2
B 1.0 m/s2
C 15 m/s2
D none of the above
Ch 2

27. ConcepTest 2.8a Acceleration I
A) yes
B) no
C) depends on the velocity
If the velocity of a car is non-zero (v ¹ 0), can the acceleration of the car be zero?

28. ConcepTest 2.8a Acceleration I
A) yes
B) no
C) depends on the velocity
If the velocity of a car is non-zero (v ¹ 0), can the acceleration of the car be zero?
Sure it can! An object moving with constant velocity has a non-zero velocity, but it has zero acceleration since the velocity is not changing.

29. Example 6: Calculate the acceleration between points A and B and B and C.
Module 3-10

30. Graphical Analysis of Linear Motion
v is slope of position vs. time graph.
a is slope of velocity vs. time graph.
Module 3-9

31. Falling Objects Giancoli, Sec 2-1, 2, 3, 4, 8
Module 4
Falling Objects
Giancoli, Sec 2-1, 2, 3, 4, 8
Giancoli, PHYSICS,6/E © Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey

32. Falling Objects
Galileo showed that for object falling from rest with no air resistance y  t2
Note that this is true when acceleration is constant
Module 4-1

33. Falling Objects Galilei showed that near the surface of the earth
at the same location
in the absence of air resistance
all objects fall with the same constant
acceleration g, the acceleration
due to gravity
g = 9.8 m/s2
Note: g is a positive number. When you define your coordinate system, if up is positive then a = - g.
Module 4-2

34. Up and Down Motion
For object that is thrown upward and returns to starting position:
assumes up is positive
velocity changes sign (direction) but acceleration does not
Velocity at top is zero
time up = time down
Velocity returning to starting position = velocity when it was released but opposite sign
Module 4-3

35. Acceleration due to Gravity
Module 4-4

36. Use quadratic equation:
(2-47) A stone is thrown vertically upward with a speed of m/s from the edge of a cliff. The distance from the bottom to the release point is 70.0 m.(a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?
Example 5
(a) UP = POSITIVE
Use quadratic equation:
Answer = 5.20 s
Module 4-5

37. (c) Find maximum height, where
Example 5
2-47 A stone is thrown vertically upward with a speed of m/s from the edge of a cliff. The distance from the bottom to the release point is 70.0 m.(a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?
(b)
(c) Find maximum height, where
Total distance =
Module 4-6

38. Example 5 (2-47) continued A stone is thrown vertically upward with a speed of m/s from the edge of a cliff. The distance from the bottom to the release point is 70.0 m. Excel Calculation—use the equation for displacement and velocity to calculate y and vy vs time.
Module 4-7

39. Problem Solving Tips
1. Read the whole problem and make sure you understand it. Then read it again.
2. Decide on the objects under study and what the time interval is.
3. Draw a diagram and choose coordinate axes.
4. Write down the known (given) quantities, and then the unknown ones that you need to find.
5. What physics applies here? Plan an approach to a solution.
Module 4-8

40. Problem Solving Tips
6. Which equations relate the known and unknown quantities? Are they valid in this situation? Solve algebraically for the unknown quantities, and check that your result is sensible (correct dimensions).
7. Calculate the solution and round it to the appropriate number of significant figures.
8. Look at the result – is it reasonable? Does it agree with a rough estimate?
9. Check the units again.
Module 4-9