velocity-time graph slope

Slope of Velocity-time Graph

The slope, or gradient, of a velocity-time graph tells us the acceleration of a body.

Suppose a body is moving at a constant velocity of 2 m/s. The acceleration of this body is zero, since there is no change in velocity. A graph of velocity against time would be a straight, horizontal line. The gradient of a horizontal line is also zero. So in this case, the gradient of the graph is equal to the acceleration.

Next suppose a body at rest starts to accelerate uniformly - its velocity increases by the same amount every second. After 5 s, the velocity increases to 2 m/s. So the acceleration = change in velocity / time taken = 2 / 5 = 0.4 m/s^s. A graph of the velocity against time would be a straight line with a slope, going through the origin. This graph passes through the point (5, 2). So its gradient is also 2 / 5 = 0.4 m/s^s, the same as the acceleration.

What if the acceleration changes? Suppose a body at rest starts accelerating at an increasing rate. So every second, the velocity increases more than in the previous second. Then the graph would be a curve with increasing slope. In this case, the acceleration would different at different times. Suppose we want to find the acceleration when the time is 3 s. The way is to draw a straight line touching the graph at 3 s, and find its gradient. We find the gradient by drawing a triangle, with the straight line as one side, and the other sides vertical and horizontal. In the graph below, the gradient = vertical / horizontal = 0.5 / 2 = 0.25 m/s^s.

To see why this is so, think of a very small time interval about t = 3 s. Suppose this time interval is b seconds. Draw the two vertical lines to the curve that contains this time interval. At the curve, draw a horizontal line in between them to show the change in velocity, a m/s. Notice that the sides a, b and the curve roughly forms a right angled triangle. So the acceleration during this time is approximately given by the change in velocity divided by the time, i.e. a/b m/s^s. This is not exact because the curved side of the triangle is not straight.

Now imagine making the time interval b very small. The triangle becomes very small, the curved side becomes very short, and also much straighter. So a/b m/s^s would give the acceleration much more accurately. It is difficult to measure such a tiny triangle. The easy way is to draw a longer, straight line touching the curve at t = 3 s. This line would be parallel to the slanted side of the triangle, so that its gradient is also equal to a/b.

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