Objectives:

• To understand graphical representations of displacement, speed, velocity and acceleration
• To be able to draw, interpret and understand displacement–time graphs, including how velocity is the gradient of such a graph
• To be able to draw, interpret and understand velocity–time graphs, including how acceleration is the gradient and displacement is the area under such a graph

Displacement-time graphs

We can represent the changing position of a moving object by drawing a displacement against time graph (d-t graph). Some of the main point to appreciate with regards to d-t graphs are:

• The gradient of the graph is equal to the object’s velocity.
• The steeper the slope the greater the velocity.
• A graph like this can also tell us if an object is moving with constant velocity or accelerating.
• The straight line graph shows that the object’s velocity is constant.
• Comparing to objects moving with constant speed, the one with a steeper gradient is moving faster.
• If the line is parallel to the x-axis, it means that the displacement is not changing, so the object is still.
• If the slope has a negative gradient, then the object is moving backwards to where it came from.
• If the line is curved, the slope is changing, which means that the velocity is changing. This means that the object has an acceleration or a deceleration which may or may not be constant.

Velocity against time graphs

A velocity against time graph (v-t graph) tells you more information about the change of velocity of an object. If the v-t graph is a curve, this may help you to understand more about its acceleration. It tells you the velocity of an object at any instant, while the slope of the graph (the gradient) tells you the acceleration at any instant. The distance traveled over any phase of the journey is equal to the area under the graph. The average speed can still be calculated as usual using equation $v = \frac{d}{t}$.

In the graph above it can be seen that the object moves with increasing speed in the first part and because the slope is a straight line, the acceleration of the object (the rate of change in velocity) is constant. Then in the second part, the object is moving with constant speed, then it increases but it doesn’t increase at the same rate at all instant, so there is an acceleration which is increasing. Then it is going at constant acceleration again and then decelerates as the velocity gradient is negative. This doesn’t mean that is travelling backwards, but just that it’s slowing down at a constant rate.