Velocity-time Graph Area



We can use the area under the velocity-time graph to tell us the displacement of a body.

Suppose a body moves at a velocity of 2 m/s for 5 s. We can find the displacement using d = vt = 2 x 5 = 10 m. Now consider the velocity-time graph for this body. It is a horizontal line. The area under this graph is a rectangle with a 5 s "long" base and a 2 m/s "tall" height. The area of this triangle is also 2 x 5 = 10 m. So in this case, the area under the graph is equal to the displacement.

What if the velocity is not uniform? It turns out that the area under the graph still gives the correct displacement. Suppose the body accelerates uniformly from rest, so that its velocity increases by the same amount every second. The velocity-time graph is a slanted, straight line through the origin. Suppose that it reaches 2 m/s after 5 s. The area under this graph is a triangle with a base of 5 s, and a height of 2 m/s. So the area = 1/2 x base x height = 1/2 x 5 x 2 = 5 m. This is also the displacement.

To see why the area is the displacement, imagine dividing up the area into many long, narrow rectangles. Pick one of these rectangles. Think of the top side of this rectangle as a horizontal graph that has constant velocity for a very short time. Then the area of a rectangle would be the displacement over this short time interval. This constant velocity is close to the actual velocity given by the slanted line just above it. If we add up the areas of all of the rectangles, we should get the approximate, total displacement. To get the actual displacement, we must make the width of each rectangle very small. Then the total area becomes the area of under the graph - in this case, the triangle.

Using the same reasoning, we can understand that the area under a velocity-time graph of any shape, curved or straight, must be equal to the displacement.




Copyright 2010 by Kai Hock. All rights reserved.
Last updated: 22 November 2010.