   Velocity–Time graphs
Introduction

In busy costal areas, shipping lanes are monitored closely to prevent accidents. The paths of ships are tracked carefully to chart courses avoiding other traffic.

The speed of a ship and the direction in which it is travelling are the key pieces of information required for shipping lanes to operate safely. Global positioning systems specify the positions of ships, and navigation aids determine their velocity. Information about the ship's velocity can be presented in graphical form for review when required.

This unit will consider the information that can be determined from a velocity versus time graph.

Velocity–time graphs
Grab the ball in the simulation of Fig.1 and drag it to the top of its range of motion. Release the ball and observe its motion and the velocity–time graph. Then answer the following questions.

Click on the figure below to interact with the model.  Figure 1.  Accelerating ball.

When the ball is falling the velocities on the graph are shown as …
• When the ball is falling the size of the
velocity
An object's velocity states both the speed and direction of motion relative to a fixed reference point.
velocity
is –
• A bouncing ball
Grab the ball in the simulation of Fig.2 and drag it to the top of its range of motion. Release the ball and observe its motion and the velocity–time graph. At first the ball falls, but then it bounces. Use the pause button to investigate how the ball's velocity changes as it bounces.

Click on the figure below to interact with the model.  Figure 2.  Bouncing ball.

At the top of each bounce the ball has a speed of …
• At the top of each bounce the speed is zero for an instant, because the ball changes direction. At the points where the speed is zero, the gradient of a distance–time graph would be zero. A tangent to the distance–time curve at these points would be a horizontal line.

From the velocity–time graph in Fig.2 we can also determine times when the ball is stationary, speeding up, or slowing down. Additionally we can specify occasions when it is rising or falling. All this information is conveyed in the velocity–time graph.

We can recognize areas when the ball is slowing down because …
• We can recognize occasions when the ball is speeding up because …
• Complete the following statements to summarize the behaviour of the ball in Fig.2.

• As the ball falls towards the floor its speed . As the ball rises towards it highest point its speed .

Just before it hits the floor its speed is maximum and it is moving . Immediately after leaving the floor, at the start of an upwards journey, its velocity is maximum but is in the direction.
• Since the graph in Fig.2 is plotting the ball's velocity, the direction in which it is travelling is important.

The ball's velocity in Fig.2 is positive when it is moving …
• The motion of the bouncing ball can be summarized as follows: When the ball is released at the top of its range of motion, it falls. Its velocity starts at zero and becomes more and more negative as it speeds up on the way down (section OA on Fig.3). Its velocity as it falls goes from 0 ms−1 to its maximum negative value just as it hits the floor (at point A on Fig.3). Contact with the floor is represented by section AB on Fig.3. After leaving the floor, it sets off upwards and once again it slows as it rises (section BC on Fig.3).
 As the ball slows down when rising, its velocity gets less and less positive until, at the highest point, it is zero and the ball is stopped momentarily (at point C on Fig.3).

 Figure 3. Graph for one bounce. Clearly the bouncing ball is one case where the velocity–time graph conveys a lot of information about speeding up, slowing down, rising, or falling.

Determining distance travelled
Set the initial speed of the car in Fig.4 to 20 ms−1 and the reaction time to 1 second. Press the button to stop the car, and note the shape of the graph as the car slows to a halt. Initially the car travels at a steady speed for a short time while the driver's braking foot reacts to the signal from his brain to stop. The length of the horizontal part of the line in the graph indicates this 'reaction time'.

 Figure 4. Stopping distances. With the initial speed of the car set to 20 ms−1 and a reaction time of 1 second, how far does the car travel while the driver's senses are reacting?
• Altering the initial speed and reaction time alters the thinking distance. If the initial speed is v and the thinking time t, then the thinking distance d is given by

Note also that the thinking distance is equal to the area under the horizontal part of the line on the graph.

Set the initial speed of the car set to 30 ms−1 and the reaction time to 1.5 seconds.
Complete the following statements.

• For seconds after the brakes are applied the car continues to travel at a speed of ms−1. The thinking distance is m.
• The thinking distance is again given by the area under the horizontal part of the line on the graph. Additionally, the distance travelled while decelerating is given by the area under the sloped part of the line. This area is a right angled triangle and its size can be calculated from its length and height.

The total distance required to stop the car is the thinking distance plus the braking distance. This is the total area under the velocity–time graph.

Determining acceleration
The
acceleration
An object's acceleration is its rate of change of velocity.
acceleration
of a moving object is defined as

A car decelerates from 30 ms−1 to rest in 5.38 s. Its acceleration can be stated as …
• In the simulation in Fig.5, the braking force used to bring the car to rest can be altered.

When the braking force in Fig.5 is increased the car …
• Set the car's initial speed in Fig.5 to 30 ms−1, the reaction time to 1 second, and the braking force to 4 kN.

Complete the following statements.

• For  s after the brakes are applied, the car continues to travel at a speed of  ms−1. When the brakes are applied, the car decelerates and comes to rest approximately  s later.
• Figure 5. Variable deceleration. Set the car's initial speed in Fig.5 to 30 ms−1, the reaction time to 1 second, and the braking force to 6 kN.

Complete the following statements.

• For second after the brakes are applied the car continues to travel at a speed of  ms−1. When the brakes are applied the car decelerates and comes to rest approximately seconds later.
• Once again, the area under the graph in this example indicates the distance travelled while decelerating. Additionally, the steepness of the line indicates how quickly the velocity is changing.

The car's
deceleration
The deceleration of an objects is a measurement of its rate of change of velocity as it slows down.
deceleration
increases when the braking force in Fig.5 is increased, and the gradient of the line in the velocity–time graph …
• The actual value for the deceleration can be found by calculating the gradient of the velocity–time graph. The decelerating car produced a velocity–time graph that was a straight line with a negative gradient.

The negative gradient of this line indicates that the car …
• The straight line nature of this line indicates that the deceleration …
• Summary

From velocity–time graphs it is possible to identify when objects are speeding up, slowing down, or travelling at a constant speed.

When an object's velocity changes from a positive value to a negative one, the direction in which the object is moving has changed.

The area under the line in a velocity–time graph gives the distance travelled.

The gradient of the line in a velocity–time graph gives the acceleration. Large accelerations are represented by steep gradients.

Exercises
 Figure 6. 1. The velocity–time graph in Fig.6 is obtained for a moving object which passes a marker at time t = 0.

• How far is the object from the marker at 2 seconds?
m   (to the nearest whole number)

How far is the object from the marker at 3 seconds?
m   (to the nearest whole number)
• 2. The motion of an object is represented by the velocity–time graph shown in Fig.7. The object is moving from left to right. Calculate the acceleration of the object during the first 3 s of its motion.
•  ms−2   (to 2 d.p.)
• Figure 7. 3. Calculate the acceleration during the final 2 s of the motion shown in Fig.7.
•  ms−2   (to 1 d.p.)
• 4. How far to the right of its starting position is the object at the end of the motion shown in Fig.7.
•  m   (to 2 d.p.)
• 5. Fig.8 and Fig.9 show the velocity–time graphs for two hot rods in a dragster race. How far does each car travel in the first 5 seconds of its motion (to the nearest whole number)?

• Car A:  m.   Car B: m.
• Figure 8. Car A. Figure 9. Car B. 6. Which car travels the greater distance in 10 seconds?
• Figure 10. 7. A constant force is used to move an object. The velocity–time graph for the motion of the object is shown in Fig.10. Which row in the following table shows the correct values for the instantaneous speed after 3 seconds and the average speed during the first 3 seconds of the motion?
• Instantaneous speed / ms−1 Average speed / ms−1 A 16.0 16.0 B 16.0 10.7 C 12.0 6.0 D 8.0 24.0 E 12.0 8.0

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