PHYSICS - Distance-time graphs.

Physics Additional Science, Distance-time graphs.

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  • Created by: Jessica
  • Created on: 12-04-10 10:41

The unit of .................. is the metre/second.

An object travelling at a steady ............ travels the same ............. every second.

The steeper the line on a distance-time graph of a moving object, the greater its ........... is.

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Speed (M/S) = distance travelled (M) / time taken

A vehicle on a motorway is travelling 1800 m in 60 seconds.

Calculate the speed of the vehicle in m/s.

Calculate how far it would travel at this speed in 300 seconds.

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SPEED (M/S) = 1800 / 60

= 30 m/s


SPEED (M/S) = 1800 / 300

= 6

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Velocity is speed in a given direction.

An object moving steadily around in a circle has a constant speed. Its direction of motion changes as it goes round therefore its velocity is NOT constant.

Two moving objects can have the same speed but different velocities.

A car travelling north at 30 m/s on a motorway has the same speed as a car travelling south at 30 m/s. But their velocities are not the same as they are moving in opposite directions.

How far apart are the cars, 10 seconds after they pass one another?

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300 metres.

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The acceleration of an object is its change of velocity per second.

What is the unit of acceleration?


We use the term deceleration when the driver breaks for example, in a car. We use the term deceleration or negative acceleration when any object slows down.

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Acceleration = m/s^2 (metre per seconf squared)

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If the velocity of an object increases by the same amount every second, its ...................... is constant.

Deceleration is when the ..................... of an object decreases.

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1. acceleration.

2. speed.

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Acceleration (m/s^2) = Change in velocity (m/s)


time taken for the change (s)

The velocity of a car increased from 8m/s to 28 m/s in 8s without change of direction. Calculate:

its change of velocity.

its acceleration.

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Acceleration (m/s^2) = change in velocity (m/s) / time taken to change (s)

Acceleration = 20 / 8

Acceleration = 2.5 m/s^2

Change in velocity (m/s) = acceleration (m/s^2) * time taken to change (s)

Change in velocity = 2.5 * 8

Change in velocity = 20 m/s

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When a velocity-time graph uses a sloped line, the section of the graph for constant velocity is flat. The lines slope is zero and so the acceleration in this section is zero therefore the object is travelling at a constant speed.

When the brakes are applied, the velocity decreases to zero and the vehicle decelerates. The slope of the line is negative in this section.

The area under a straight line represents constant velocity. Therefore, to find out the distance travelled, use the velocity * time.

The area under a sloped line represents deceleration. You can work out the distance travelled in this section by using the area of the section under the sloped line. This will be 1/2 * height * base of the triangle.

The slope of the line on a velocity-time graph represents acceleration.

The area under the line on a velocity-time graph represents distance travelled.

Would the total distance travelled be greater or smaller if the deceleration had taken longer?

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For an object with a changing speed the distance-time graph is not a straight line. They are often curved lines.

The slope on a distance-time graph represents speed.

The slope on a velocity-time graph represents acceleration.

The area under the line on a velocity-time graph represents the distance travelled.

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