1. Module 1.1 – Position, Distance, Displacement
This module is an introduction to some basic kinematics terminology that will be important throughout the entire unit, as well as in future units.

2. Scalars and Vectors
Scalar quantities - do not have a direction associated with them. They have a magnitude (size) only.
Vector quantities have two parts to them:
A magnitude of the quantity, and
A direction

3. Position
Position ( ) - vector quantity. A position vector locates an object. Must indicate
Magnitude
Direction
Origin

4. Check Your Learning Which of the following is a vector quantity?
2.0 kg
34 km/h
34 km/h east
3.2 s
C – this is the only measurement with both magnitude and direction
Which of the following describes the position of an object?
45 km
45 km east
45 km east of the flagpole
Since position must indicate the location of an object, it must include a magnitude, a direction, and a reference point; therefore, answer c is the only one that correctly describes a position.
A person leaves their house and walks 2.0 km east, takes a break, and then walks another 4.2 km east. What is their position?
Their position is 6.2 km east of their house.

5. Distance and Displacement
Distance ( ) - scalar quantity, refers to the total movement of an object without any regard for direction. Distance is what an odometer in a car records.
Displacement ( ) is a vector and refers to the change in position of an object:

6. Example 1
A runner in a race is at a position of 43.2 m to the right of the starting line when you first notice him. You turn your head briefly, and then look back. He is now 82.6 m to the right of the starting line. What was his displacement?

7. Example 1 Answer
The displacement is 39.4 m to the right.

8. Example 2 A person walks 5.0 km east, and then 4.0 km east.
What distance has the person travelled?
What is the person’s displacement?

9. Example 2 Answer
This time, there are two individual displacements given instead of the intial and final positions
Since distance is not a vector, no direction should be provided when describing the distance.

10. Example 2 Answer
This time, there are two individual displacements given instead of the intial and final positions .
Since displacement is a vector, it must also have a direction stated. The displacement is therefore

11. Example 3 A person walks 5.0 km east, and then 4.0 km west.
What distance has the person travelled?
What is the person’s displacement?

12. Example 3 Answer
As can be seen in the diagram, the total distance travelled by the person is still
Remember that displacement compares the person’s final position with their initial position.
Since the displacement is positive, this tells us that the displacement is 1.0 km east.

13. Check Your Learning
A person walks 850 m north, 320 m south, and then 130 m north.
What is her distance walked?
Since direction is not important when calculating distance, the total distance is found just by adding each of the individual distances.

14. Check Your Learning
What is her displacement?
When calculating the displacement, the directions are important. To do this, we can consider all displacements north to be positive and all directions south to be negative.
Since the answer is positive, the displacement is 660 m north.

15. Check Your Learning
A runner completes a race by running around a 200. m circumference circular track 3 times.
What distance has he travelled?
Total distance is
What is his displacement?
Since the runner ran 3 complete laps around the circular track, he arrived back at his original position. Since the displacement is the change in position, the runner’s displacement is

16. Position-time Graphs Position (km) Time (s) 3 1 6 2 9 12 4 15 5 18 213 1 6 2 9 12 4 15 5 18 21 7 24 8

17. Check Your Learning
Plot a position-time graph of the following data and answer the following questions:
Position (m)
Time (s) 10 20 2 30 4 40 6 50 8
What is the position at 1.0 s?
15m
What is the position at 5.0 s?
35m
What is the displacement between 1.0 and 5.0 s?

18. Module 1.2 – Speed and Velocity
This module introduces the concepts of average speed and velocity. It is important to master these concepts before going on to study accelerated motion.

19. Definitions Speed is defined as the rate of change of distance.
Velocity is defined as the rate of change of displacement.

20. Example 1
A person runs a 100. m race in 11.3 s. What was their average speed?
The only object of interest in this problem is the person running.
We must write down all of our information:

21. Example 1 Answer
The units in the answer (m/s) can be obtained from the equation using dimensional analysis

22. Example 2
A person drives 35 km East in 35 minutes, then 52 km West in 48 minutes.
What was the person’s average speed in km/h?
What was the person’s average velocity in km/h?

23. Example 2 Answer
Writing down our information,

24. Example 2 Answer
Writing down our information (and using the right as our positive direction), we have
The average velocity is
12 km/h west.

25. Unit Conversion
A common conversion in physics is from km/h to m/s (or m/s to km/h).
Conversion factor is
Divide by 3.6 to convert from km/h to m/s
Multiply by 3.6 to convert from m/s to km/h

26. Check Your Learning
If you are driving 90.0 km/h and you look to the side for 2.0 s, how far do you travel during this inattentive period?

27. Check Your Learning
A person jogs 1.4 km east in 12 minutes, and then 3.5 km west in 21 minutes.
What was their average speed?

28. Check Your Learning What was their average velocity?
The average velocity was 3.8 km/h west.

29. Example 3
A person runs at a speed of 4.0 m/s for 6.3 minutes, and then 3.0 m/s for the last 8.2 minutes. What was their average speed?

30. Check Your Learning
An airplane travels 1800 km at a speed of km/h. It then encounters a headwind that slows it down to 850 km/h for the next 2300 km. What was the average speed of the plane for this trip?

31. Position-time Graphs represents the slope of the graph
position (m)
time (s) 3 5 1 7 2 9 11 4 13 15 6
represents the slope of the graph
A linear position-time graph represents a constant velocity.

32. Example
Using the previous graph, calculate the velocity by finding the slope. or

33. Check Your Learning
Answer the following questions based on the position-time graph below:

34. Check Your Learning When is the object at rest? Between 4.0 and 10.0 s
When is the object moving forward?
Between 0 and 4.0 s
When is the object moving backward (toward its starting point)?
Between 10.0 and 12.0 s

35. Check Your Learning What is the object’s velocity between 0 and 4.0 s?

36. Instantaneous Velocity
Describe the motion in the following graph:
Calculate the average speed for the first 4 seconds
How fast is the object moving at 4.0 seconds?

37. Instantaneous Velocity
Could find the average velocity for an interval around 4.0 s …
How could we make this estimate more precise?
As the interval becomes smaller and smaller, the line becomes a tangent line

38. Tangent Line
To be more precise, this interval length should approach zero …
Slope of the tangent line approximates the slope of the graph at the point of interest
It should be drawn so that it just touches the curve at the point of interest

39. Example 1 Find the displacement of the object in the graph.
Since velocity is constant
Actually calculated area under graph!

40. Example 2
Find the displacement of the object in the graph (not constant velocity).
Calculating area of triangle gives

41. Area=Displacment
The area under a velocity-time graph gives the displacement of an object.

42. Check Your Learning
Answer the following questions based on the velocity-time graph below:

43. Check Your Learning When is the object moving at a constant speed?
Between 4.0 s and 10.0 s
When is the object moving forward?
The whole time – the velocity is positive for the entire 12.0 s
When is the object slowing down?
Between 10.0 and 12.0 s
What is the displacement between 0 and 4.0 s?
Remember, displacement is the area of a velocity-time graph and the shape of the graph between 0.0 and 4.0 s is a triangle, whose area is given by

44. Relative Velocity
All velocities are relative to some frame of reference (usually the Earth)
Some problems can be simplified by changing our frame of reference.

45. Example
A car traveling 90. km/h is 100. m behind a truck traveling 50. km/h. How long will it take the car to reach the truck?

46. Example – Answer
Instead of looking at each object separately (using their speeds relative to the earth) we will change our reference frame to be the truck.
The car is going 40. km/h faster than the truck. Assume truck is not moving:

47. Calculating Relative Velocity
If both velocities are originally given in the same frame of reference, relative velocity is
Where refers to the velocity of object 1 relative to object 2
Consider now the case where the car and truck are moving toward each other with the same speeds as in the example
Relative velocity is now larger than either individual velocity

48. Check Your Learning
Train A leaves from New York on its way to Chicago traveling at 130 km/h. Train B leaves Chicago on its way to New York traveling at 180 km/h. The distance from New York to Chicago is 1200 km.
What is the relative velocity of the trains?
Take the train traveling from New York to be moving in the positive direction:

49. Check Your Learning How long does it take the trains to meet?
How far from New York do they meet?

50. Module 1.3 Acceleration
In the real world, objects seldom move only with constant velocity and we are often interested in more than just average velocities. This module extends the study of kinematics to all motion involving constant acceleration.

51. Definition
Acceleration ( ) tells us how quickly the velocity is changing and is defined as the rate of change of velocity:
Units:
and

52. Example
A car accelerates from rest to a velocity of 80.0 km/h in 6.3 s. What was the average acceleration?

53. Check Your Learning
Answer each of the following questions, providing an example for each.
If the velocity of an object is zero, does that mean that the acceleration is zero?
No, it does not. Consider the case of a car at rest. When you step on the gas peddle, the car starts accelerating. If the acceleration were zero, the velocity would not change and the car would never move!
If the acceleration of an object is zero, does that mean that the velocity is zero?
No, it does not. Any object moving at a constant velocity (such as a car driving in a straight line at a constant speed) has an acceleration of zero.

54. Check Your Learning
A car traveling at 75 km/h slows down to 25 km/h in 4.3 s. What was the acceleration?
Writing down all of our information (and using the direction that the car is moving as positive), we have
Note that the acceleration of the car is negative, meaning that it is in the direction opposite to that in which it is moving (and has the opposite sign of the car’s initial velocity). In other words, every second the car will be losing 3.2 m/s of its velocity.

55. Acceleration and Velocity-time Graphs
velocity (m/s)
time (s) 3 5 1 7 2 9 11 4 13 15 6
represents the slope of the graph
A linear velocity-time graph therefore represents a constant acceleration.

56. Example
Using the previous graph, calculate the acceleration by finding the slope or

57. Check Your Learning
Calculate the constant acceleration of the object shown in the velocity-time graph below:

58. First Displacement Equation
Consider the following velocity-time graph with constant acceleration:
Area=displacement
Draw a line across the graph half way between vi and vf
Area under this horizontal line should be the same as the area under the original line, since the triangle cut off by the line (A) has the same area as the triangle added (B)

59. First Displacement Equation
Horizontal line is drawn halfway between vi and vf so its value will be
Which is average velocity
Area under this line therefore represents the displacement or

60. Example
A car accelerates from 10.0 km/h to 45.0 km/h in 13.0 s. How far has it traveled?

61. Check Your Learning
A car travels 125 m in 9.3 s while accelerating from an initial speed of 33.0 km/h. What is the car’s final speed?

62. Second Displacement Equation
Consider the same velocity-time graph with constant acceleration that we looked at before:
We can break the area under the line into 2 shapes, a triangle and a rectangle
The total displacement can then be found by finding the area of A and B and adding them together

63. Second Displacement Equation
But we know that

64. Example
A person running at 2.3 m/s starts accelerating. While accelerating, they travel 4.6 m in 1.6 s. What was their acceleration?

65. Check Your Learning
A car starts from rest and accelerates at 2.3 m/s2, covering a distance of 76 m. How long did this take?

66. Quadratic Formula For any quadratic equation of the form
Where a,b,c are known coefficients, the variable t can be solved for using the following equation:

67. Example
A car has an initial speed of 25.0 km/h, and travels 44.6 m while accelerating at 1.6 m/s2. How long does this take?

68. Example so

69. Third Displacement Equation
We now want an equation that does not have time as a variable. If we start with
We must replace the variable t with another expression:
Combining these equations gives or

70. Example
A cyclist travels a distance of 54.0 m while accelerating from 2.0 m/s to 6.4 m/s. What was her acceleration?

71. Check Your Learning
A car with an acceleration of 2.0 m/s2 accelerates from 32 km/h to 44 km/h. How far does the car travel while doing this?

72. Problem Solving Strategies
It is important to note that physics is not simply a collection of equations to be memorized. Simply searching for an equation that might work will often lead to a wrong answer. It is important to understand the physics of a situation before trying to solve it.
Read and reread the problem to make sure that you understand the situation. Do not just pick out numbers from the problem.
Decide what object (or objects) that you are going to focus on in the problem.
Decide how many sequential events there are in the problem. For example, if a car accelerates for a period of time and then moves at a constant speed this can be considered to be two separate events in the problem.
Draw a diagram or picture of the situation. Choose a direction (or directions) as positive. For example, you may choose the positive x-axis (or right) and the positive y-axis (or up) as the positive directions.

73. Problem Solving Strategies
Write down all of the given information
If there is more than one object in the problem (as identified in step 2), write down the information for each object separately.
If there is more than one logical part (or event) to the problem (as identified in step 3), write down the information separately for each part.
Check all of the units for the quantities that you wrote down – do they match up? For example, if the velocity is in m/s and the distance is in km then the distance must be converted into m.
Determine what the question is asking you to find – what is the unknown quantity?
What physics is involved in the problem? Is it a kinematics problem? Are there forces involved? Is it a one dimensional or a two dimensional problem? Is it a circular motion problem? Answering this question will help determine what other steps may be needed; for example, if the problem involves forces it will be necessary to draw a free body diagram.

74. Problem Solving Strategies
Review the list of relevant equations (as determined by the previous step) to determine what equations may possibly be used to calculate the unknown quantity. If you find an applicable equation that involves only known quantities and the desired unknown quantity, solve the problem using that equation. In many cases, several sequential calculations using a combination of equations may be required.
While extra significant digits can (and probably should be) carried through intermediate steps, the final answer should be rounded to the appropriate number of significant digits.
Is the result reasonable? Does it make sense? Double check to make sure that the units balance; in other words, that the units used in the problem give the proper units in the answer.

75. Acceleration due to Gravity
Free fall refers to an object that is only being acted on by the force of gravity and is therefore undergoing a constant acceleration downward.
All objects near the surface of the Earth (as long as we are ignoring air resistance) fall with the same acceleration (g)

76. Example
A rocket is released from the ground with an upward velocity of 23.0 m/s.
How high does it go?
How long is it in the air?
What is the velocity of the rocket upon its return to the ground?

77. Example – Answer We will use the up direction as positive:
Since we are looking for the height of the object, we know that the final velocity will be zero and the acceleration is 9.80 m/s2 downward.
We will use the up direction as positive:

78. Example – Answer
It is necessary to break the problem up into two parts – the trip up and the trip down. We will still continue to use up as positive
The total time in the air is therefore

79. Example – Answer
Using the trip down again (and still using the up direction as positive),
Notice that there are two possible mathematical answers here; since it is known that the rocket is moving down when it returns to the ground, and we chose up as positive, the correct answer is

80. Things to Remember
The instantaneous vertical speed of any object at the top of its path is zero.
The time it takes an object to go up is the same amount of time that it takes to come back down (ignoring air resistance and assuming it returns to the same level as it left).
The speed at which an object returns to its starting point is the same (although the sign of the velocity is opposite) as that at which it left.
The sign that we give to the acceleration depends only on what direction we chose as positive.

81. Check Your Learning
A ball is thrown in the air with an initial speed of 4.0 m/s.
What is the height of the ball when the velocity is 1.5 m/s upward?
Writing down our information, and using up as positive, we have

82. Check Your Learning
What is the acceleration of the ball at the highest point?
The acceleration of the ball is still Just because the ball’s velocity is zero at the highest point does not mean that the acceleration changes. The acceleration due to gravity remains constant.

83. Unit Summary Module 1.1 – Position and Displacement
In this module you learned
The difference between a scalar and a vector
How to identify the position of an object
The difference between, and how to calculate, distance and displacement
How to plot, and obtain information from, a position-time graph.

84. Unit Summary Module 1.2 – Speed and Velocity
In this module you learned
The difference between speed and velocity and how to calculate each in a variety of situations.
That the slope of a position-time graph represents an object’s velocity.
That the area of a velocity-time graph represents an object’s displacement.
How to use frames of reference and relative velocity to simplify problems.

85. Unit Summary Module 1.3 – Acceleration In this module you learned
How to calculate acceleration and obtain acceleration from a velocity-time graph
How to use the kinematics equations derived in this section to solve a variety of problems
That the acceleration due to gravity is 9.80 m/s2 and that the kinematics equations can be used to solve problems involving the acceleration due to gravity.

86. Unit Summary