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Description of motion

Module by: Free High School Science Texts Project. E-mail the author

Description of Motion

The purpose of this chapter is to describe motion, and now that we understand the definitions of displacement, distance, velocity, speed and acceleration, we are ready to start using these ideas to describe how an object is moving. There are many ways of describing motion:

  1. words
  2. diagrams
  3. graphs

These methods will be described in this section.

We will consider three types of motion: when the object is not moving (stationary object), when the object is moving at a constant velocity (uniform motion) and when the object is moving at a constant acceleration (motion at constant acceleration).

Stationary Object

The simplest motion that we can come across is that of a stationary object. A stationary object does not move and so its position does not change, for as long as it is standing still. An example of this situation is when someone is waiting for something without moving. The person remains in the same position.

Lesedi is waiting for a taxi. He is standing two metres from a stop street at tt = 0 s. After one minute, at tt = 60 ss, he is still 2 metres from the stop street and after two minutes, at tt = 120 ss, also 2 metres from the stop street. His position has not changed. His displacement is zero (because his position is the same), his velocity is zero (because his displacement is zero) and his acceleration is also zero (because his velocity is not changing).

Figure 1
Figure 1 (PG10C2_020.png)

We can now draw graphs of position vs. time (xx vs. tt), velocity vs. time (vv vs. tt) and acceleration vs. time (aa vs. tt) for a stationary object. The graphs are shown in Figure 2. Lesedi's position is 2 metres from the stop street. If the stop street is taken as the reference point, his position remains at 2 metres for 120 seconds. The graph is a horizontal line at 2 m. The velocity and acceleration graphs are also shown. They are both horizontal lines on the xx-axis. Since his position is not changing, his velocity is 0 m·s-10m·s-1 and since velocity is not changing, acceleration is 0 m·s-20m·s-2.

Figure 2: Graphs for a stationary object (a) position vs. time (b) velocity vs. time (c) acceleration vs. time.
Figure 2 (PG10C2_021.png)
Definition 1: Gradient

The gradient of a line can be calculated by dividing the change in the yy-value by the change in the xx-value.

m = ΔyΔxΔyΔx

Since we know that velocity is the rate of change of position, we can confirm the value for the velocity vs. time graph, by calculating the gradient of the xx vs. tt graph.

Tip:

The gradient of a position vs. time graph gives the velocity.

If we calculate the gradient of the xx vs. tt graph for a stationary object we get:

v = Δ x Δ t = x f - x i t f - t i = 2 m - 2 m 120 s - 60 s ( initial position = final position ) = 0 m · s - 1 ( for the time that Lesedi is stationary ) v = Δ x Δ t = x f - x i t f - t i = 2 m - 2 m 120 s - 60 s ( initial position = final position ) = 0 m · s - 1 ( for the time that Lesedi is stationary )
(1)

Similarly, we can confirm the value of the acceleration by calculating the gradient of the velocity vs. time graph.

Tip:

The gradient of a velocity vs. time graph gives the acceleration.

If we calculate the gradient of the vv vs. tt graph for a stationary object we get:

a = Δ v Δ t = v f - v i t f - t i = 0 m · s - 1 - 0 m · s - 1 120 s - 60 s = 0 m · s - 2 a = Δ v Δ t = v f - v i t f - t i = 0 m · s - 1 - 0 m · s - 1 120 s - 60 s = 0 m · s - 2
(2)

Additionally, because the velocity vs. time graph is related to the position vs. time graph, we can use the area under the velocity vs. time graph to calculate the displacement of an object.

Tip:

The area under the velocity vs. time graph gives the displacement.

The displacement of the object is given by the area under the graph, which is 0m0m. This is obvious, because the object is not moving.

Motion at Constant Velocity

Motion at a constant velocity or uniform motion means that the position of the object is changing at the same rate.

Assume that Lesedi takes 100s100s to walk the 100m100m to the taxi-stop every morning. If we assume that Lesedi's house is the origin, then Lesedi's velocity is:

v = Δ x Δ t = x f - x i t f - t i = 100 m - 0 m 100 s - 0 s = 1 m · s - 1 v = Δ x Δ t = x f - x i t f - t i = 100 m - 0 m 100 s - 0 s = 1 m · s - 1
(3)

Lesedi's velocity is 1 m··s-1-1. This means that he walked 1m1m in the first second, another metre in the second second, and another in the third second, and so on. For example, after 50s50s he will be 50m50m from home. His position increases by 1m1m every 1s1s. A diagram of Lesedi's position is shown in Figure 3.

Figure 3: Diagram showing Lesedi's motion at a constant velocity of 1 m··s-1-1
Figure 3 (PG10C2_022.png)

We can now draw graphs of position vs.time (xx vs. tt), velocity vs.time (vv vs. tt) and acceleration vs.time (aa vs. tt) for Lesedi moving at a constant velocity. The graphs are shown in Figure 4.

Figure 4: Graphs for motion at constant velocity (a) position vs. time (b) velocity vs. time (c) acceleration vs. time. The area of the shaded portion in the vv vs. tt graph corresponds to the object's displacement.
Figure 4 (PG10C2_023.png)

In the evening Lesedi walks 100m100m from the bus stop to his house in 100s100s. Assume that Lesedi's house is the origin. The following graphs can be drawn to describe the motion.

Figure 5: Graphs for motion with a constant negative velocity (a) position vs. time (b) velocity vs. time (c) acceleration vs. time. The area of the shaded portion in the vv vs.tt graph corresponds to the object's displacement.
Figure 5 (PG10C2_024.png)

We see that the vv vs. tt graph is a horisontal line. If the velocity vs. time graph is a horisontal line, it means that the velocity is constant (not changing). Motion at a constant velocity is known as uniform motion.

We can use the xx vs. tt to calculate the velocity by finding the gradient of the line.

v = Δ x Δ t = x f - x i t f - t i = 0 m - 100 m 100 s - 0 s = - 1 m · s - 1 v = Δ x Δ t = x f - x i t f - t i = 0 m - 100 m 100 s - 0 s = - 1 m · s - 1
(4)

Lesedi has a velocity of -1 m·s-1-1m·s-1, or 1 m·s-11m·s-1 towards his house. You will notice that the vv vs. tt graph is a horisontal line corresponding to a velocity of -1 m·s-1-1m·s-1. The horizontal line means that the velocity stays the same (remains constant) during the motion. This is uniform velocity.

We can use the vv vs. tt to calculate the acceleration by finding the gradient of the line.

a = Δ v Δ t = v f - v i t f - t i = 1 m · s - 1 - 1 m · s - 1 100 s - 0 s = 0 m · s - 2 a = Δ v Δ t = v f - v i t f - t i = 1 m · s - 1 - 1 m · s - 1 100 s - 0 s = 0 m · s - 2
(5)

Lesedi has an acceleration of 0 m·s-20m·s-2. You will notice that the graph of aa vs.tt is a horisontal line corresponding to an acceleration value of 0 m·s-20m·s-2. There is no acceleration during the motion because his velocity does not change.

We can use the vv vs. tt to calculate the displacement by finding the area under the graph.

v = Area under graph = × b = 100 × ( - 1 ) = - 100 m v = Area under graph = × b = 100 × ( - 1 ) = - 100 m
(6)

This means that Lesedi has a displacement of 100m100m towards his house.

Velocity and acceleration

  1. Use the graphs in Figure 4 to calculate each of the following:
    1. Calculate Lesedi's velocity between 50s50s and 100s100s using the xx vs. tt graph. Hint: Find the gradient of the line.
    2. Calculate Lesedi's acceleration during the whole motion using the vv vs. tt graph.
    3. Calculate Lesedi's displacement during the whole motion using the vv vs. tt graph.
    Click here for the solution
  2. Thandi takes 200s200s to walk 100m100m to the bus stop every morning. In the evening Thandi takes 200s200s to walk 100m100m from the bus stop to her home.
    1. Draw a graph of Thandi's position as a function of time for the morning (assuming that Thandi's home is the reference point). Use the gradient of the xx vs. tt graph to draw the graph of velocity vs. time. Use the gradient of the vv vs. tt graph to draw the graph of acceleration vs. time.
    2. Draw a graph of Thandi's position as a function of time for the evening (assuming that Thandi's home is the origin). Use the gradient of the xx vs. tt graph to draw the graph of velocity vs. time. Use the gradient of the vv vs. tt graph to draw the graph of acceleration vs. time.
    3. Discuss the differences between the two sets of graphs in questions 2 and 3.
    Click here for the solution

Experiment : Motion at constant velocity

Aim:

To measure the position and time during motion at constant velocity and determine the average velocity as the gradient of a “Position vs. Time" graph.

Apparatus:

A battery operated toy car, stopwatch, meter stick or measuring tape.

Method

  1. Work with a friend. Copy the table below into your workbook.
  2. Complete the table by timing the car as it travels each distance.
  3. Time the car twice for each distance and take the average value as your accepted time.
  4. Use the distance and average time values to plot a graph of “Distance vs. Time" onto graph paper. Stick the graph paper into your workbook. (Remember that “A vs. B" always means “y vs. x").
  5. Insert all axis labels and units onto your graph.
  6. Draw the best straight line through your data points.
  7. Find the gradient of the straight line. This is the average velocity.

Results:

Table 1
Distance (m) Time (s)
  1 2 Ave.
0      
0,5      
1,0      
1,5      
2,0      
2,5      
3,0      

Conclusions:

Answer the following questions in your workbook:

  1. Did the car travel with a constant velocity?
  2. How can you tell by looking at the “Distance vs. Time" graph if the velocity is constant?
  3. How would the “Distance vs. Time" look for a car with a faster velocity?
  4. How would the “Distance vs. Time" look for a car with a slower velocity?

Motion at Constant Acceleration

The final situation we will be studying is motion at constant acceleration. We know that acceleration is the rate of change of velocity. So, if we have a constant acceleration, this means that the velocity changes at a constant rate.

Let's look at our first example of Lesedi waiting at the taxi stop again. A taxi arrived and Lesedi got in. The taxi stopped at the stop street and then accelerated as follows: After 1s1s the taxi covered a distance of 2,5m2,5m, after 2s2s it covered 10m10m, after 3s3s it covered 22,5m22,5m and after 4s4s it covered 40m40m. The taxi is covering a larger distance every second. This means that it is accelerating.

Figure 6
Figure 6 (PG10C2_025.png)

To calculate the velocity of the taxi you need to calculate the gradient of the line at each second:

v 1 s = Δ x Δ t = x f - x i t f - t i = 5 m - 0 m 1 , 5 s - 0 , 5 s = 5 m · s - 1 v 1 s = Δ x Δ t = x f - x i t f - t i = 5 m - 0 m 1 , 5 s - 0 , 5 s = 5 m · s - 1
(7)
v 2 s = Δ x Δ t = x f - x i t f - t i = 15 m - 5 m 2 , 5 s - 1 , 5 s = 10 m · s - 1 v 2 s = Δ x Δ t = x f - x i t f - t i = 15 m - 5 m 2 , 5 s - 1 , 5 s = 10 m · s - 1
(8)
v 3 s = Δ x Δ t = x f - x i t f - t i = 30 m - 15 m 3 , 5 s - 2 , 5 s = 15 m · s - 1 v 3 s = Δ x Δ t = x f - x i t f - t i = 30 m - 15 m 3 , 5 s - 2 , 5 s = 15 m · s - 1
(9)

From these velocities, we can draw the velocity-time graph which forms a straight line.

The acceleration is the gradient of the vv vs. tt graph and can be calculated as follows:

a = Δ v Δ t = v f - v i t f - t i = 15 m · s - 1 - 5 m · s - 1 3 s - 1 s = 5 m · s - 2 a = Δ v Δ t = v f - v i t f - t i = 15 m · s - 1 - 5 m · s - 1 3 s - 1 s = 5 m · s - 2
(10)

The acceleration does not change during the motion (the gradient stays constant). This is motion at constant or uniform acceleration.

The graphs for this situation are shown in Figure 7.

Figure 7: Graphs for motion with a constant acceleration (a) position vs. time (b) velocity vs. time (c) acceleration vs. time.
Figure 7 (PG10C2_026.png)

Velocity from Acceleration vs. Time Graphs

Just as we used velocity vs. time graphs to find displacement, we can use acceleration vs. time graphs to find the velocity of an object at a given moment in time. We simply calculate the area under the acceleration vs. time graph, at a given time. In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the shaded portion.

v = area of rectangle = a × Δ t = 5 m · s - 2 × 2 s = 10 m · s - 1 v = area of rectangle = a × Δ t = 5 m · s - 2 × 2 s = 10 m · s - 1
(11)

The velocity of the object at t=2st=2s is therefore 10 m·s-110m·s-1. This corresponds with the values obtained in Figure 7.

Summary of Graphs

The relation between graphs of position, velocity and acceleration as functions of time is summarised in Figure 8.

Figure 8: Position-time, velocity-time and acceleration-time graphs.
Figure 8 (PG10C2_029_1.png)

Tip:

Often you will be required to describe the motion of an object that is presented as a graph of either position, velocity or acceleration as functions of time. The description of the motion represented by a graph should include the following (where possible):

  1. whether the object is moving in the positive or negative direction
  2. whether the object is at rest, moving at constant velocity or moving at constant positive acceleration (speeding up) or constant negative acceleration (slowing down)

You will also often be required to draw graphs based on a description of the motion in words or from a diagram. Remember that these are just different methods of presenting the same information. If you keep in mind the general shapes of the graphs for the different types of motion, there should not be any difficulty with explaining what is happening.

Experiment: Position versus time using a ticker timer

Aim:

To measure the position and time during motion and to use that data to plot a “Position vs. Time" graph.

Apparatus:

Trolley, ticker tape apparatus, tape, graph paper, ruler, ramp

Method:

  1. Work with a friend. Copy the table below into your workbook.
  2. Attach a length of tape to the trolley.
  3. Run the other end of the tape through the ticker timer.
  4. Start the ticker timer going and roll the trolley down the ramp.
  5. Repeat steps 1 - 3.
  6. On each piece of tape, measure the distance between successive dots. Note these distances in the table below.
  7. Use the frequency of the ticker timer to work out the time intervals between successive dots. Note these times in the table below,
  8. Work out the average values for distance and time.
  9. Use the average distance and average time values to plot a graph of “Distance vs. Time" onto graph paper. Stick the graph paper into your workbook. (Remember that “A vs. B" always means “y vs. x").
  10. Insert all axis labels and units onto your graph.
  11. Draw the best straight line through your data points.

Results:

Table 2
Distance (m) Time (s)
1 2 Ave. 1 2 Ave.
           
           
           
           
           
           
           

Discussion:

Describe the motion of the trolley down the ramp.

Worked Examples

The worked examples in this section demonstrate the types of questions that can be asked about graphs.

Exercise 1: Description of motion based on a position-time graph

The position vs. time graph for the motion of a car is given below. Draw the corresponding velocity vs. time and acceleration vs. time graphs, and then describe the motion of the car.

Figure 9
Figure 9 (PG10C2_038.png)

Exercise 2: Calculations from a velocity vs. time graph

The velocity vs. time graph of a truck is plotted below. Calculate the distance and displacement of the truck after 15 seconds.

Figure 12
Figure 12 (PG10C2_041.png)

Exercise 3: Velocity from a position vs. time graph

The position vs. time graph below describes the motion of an athlete.

  1. What is the velocity of the athlete during the first 4 seconds?
  2. What is the velocity of the athlete from t=4t=4 s to t=7t=7 s?

Figure 13
Figure 13 (PG10C2_042.png)

Exercise 4: Drawing a vv vs. tt graph from an aa vs. tt graph

The acceleration vs. time graph for a car starting from rest, is given below. Calculate the velocity of the car and hence draw the velocity vs. time graph.

Figure 14
Figure 14 (PG10C2_043.png)

Graphs

  1. A car is parked 10m10m from home for 10 minutes. Draw a displacement-time, velocity-time and acceleration-time graphs for the motion. Label all the axes.
    Click here for the solution.
  2. A bus travels at a constant velocity of 12 m·s-112m·s-1for 6 seconds. Draw the displacement-time, velocity-time and acceleration-time graph for the motion. Label all the axes.
    Click here for the solution.
  3. An athlete runs with a constant acceleration of 1 m·s-21m·s-2 for 4s4s. Draw the acceleration-time, velocity-time and displacement time graphs for the motion. Accurate values are only needed for the acceleration-time and velocity-time graphs.
    Click here for the solution.
  4. The following velocity-time graph describes the motion of a car. Draw the displacement-time graph and the acceleration-time graph and explain the motion of the car according to the three graphs.
    Figure 16
    Figure 16 (PG10C2_027.png)
    Click here for the solution.
  5. The following velocity-time graph describes the motion of a truck. Draw the displacement-time graph and the acceleration-time graph and explain the motion of the truck according to the three graphs.
    Figure 17
    Figure 17 (PG10C2_028.png)
    Click here for the solution.

This simulation allows you the opportunity to plot graphs of motion and to see how the graphs of motion change when you move the man.

Figure 18
Figure 18 (motion.png)
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