Velocity-time graph of a particle moving in a straight line is shown in figure. Plot the corresponding displacement-time graph of the particle if at time `t=0,` displacement `s=0.`

Explanation

Displacement = area under velocity -time graph.
Hence, `s_(OA) = (1)/(2) xx 2xx 10 = 10`m
`" " s_(AB) = 2xx 10 = 20`m
or `" " s_(OAB) = 10 + 20 = 30 `m
`" " s _(BC) = 2 xx((10+ 20)/(2)) = 30` m
or `" " s_(OABC) = 30 + 30 = 60 `m
and `" " s _(CD) = (1)/(2)(2xx 20) = 20`m
or `" " s_(OABCD) = 60+ 20 = 80 ` m
Between 0 to 2s and 4 to 6 s motion is acceleratecd, hence displacement-time graph is a parabola. Between 2 to 4 s motion is uniform, so displacement-time graph will be a straight line. Between 6 to 8 s motion is decelerated hence displacement-time graph is again a parabola but inverted in shape. At the end of 8s velocity is zero, therefore, slope of displacement-time graph should be zero. The corresponding graph is shown in the figure.