Given below is the velocity-time graph for a runner who accelerates initially for 2 s, then moves with uniform speed for 8 s , then decelerating for another 2 s and finally moving with uniform speed for 4 s.


It is important to note that the length of each box in the figure above is 2 seconds along the horizontal direction and 2 m/s along the vertical.
The displacement of the runner is given by the area under the velocity-time curve but the area under the magnitude of velocity and time graph gives the distance. This can be seen from the definition of distance in integral form given as:

To make it easy, divide the area into squares, rectangles and triangles (as shown in the figure above).
From the figure above, we have that the area under the curve is the sum of area of triangle ABC, rectangle BDEC, triangle DOF, rectangle OFGE and square FHIG respectively.
Now area of triangle ABC (say A1) is:


Area of rectangle BDEC (say A2) is:

Area of triangle DOF (say A3)is:

Area of rectangle OFGE (say A4)is:

Area of square FHIG (say A5)is:

The total area (say s) under the curve is:


Therefore the runner has gone 100 m in 16 seconds.