In other activities, we have looked at distance-time and velocity-time graphs where the lines have been straight. In these questions, the gradient (or rate of change) of the diagonal lines has given the time (for a distance-time graph) and the acceleration (for a velocity-time graph).

In real life, this is a false situation as objects rarely travel or accelerate at a constant rate.

It is more likely that if you sketched either of these graphs, you would get a curve.

So this begs the question, how do you find the gradient when you have a curve?

**Average and instantaneous**

These questions can take one of two forms.

The following details are referring to velocity-time graphs, but the same rules apply for distance-time, just replace the word acceleration with the word velocity.

You are asked for the **average **acceleration. In these questions, you are finding the acceleration between two points. All you have to do is draw a straight line between the two points and find the gradient of that line.

You are asked for the **acceleration at a specific time - ** this is called the **instantaneous** acceleration.

For these questions you are finding the gradient at a given point on the graph. Without going into calculus (this is still an A-level only topic), all you need to know is that the gradient of a curve at a point is the same as the gradient of the tangent at that point.

**Example:**

The velocity of a Formula 1 car over the first 5 seconds of a race is shown on this velocity-time graph:

**Find the average acceleration over the first 3 seconds.**

Average means that we are looking for the acceleration between t = 0 and t = 3. We need to draw a line between these two points on the graph and find the gradient.

**This gives an average acceleration of 57.3 m/s ^{2}**

** Find the acceleration 2 seconds after the start of the race.**

In this question, we are asked for the acceleration at a specific point - this means that we have an instantaneous rate of change. We need to draw a tangent at t = 2 and find the gradient at this point.

This gives an **instantaneous acceleration of 23.1 m/s ^{2}**