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_(DA)=(1)/(2)xxxx10=10m`
`s_(AB)=2xx10=20m`
or `s_(OAB)=10+20=30m`
`s_(BC)=2xx((10+20)/(2))=30m`
or `s_(OABC)=30+30=60m`
and `s_(CD)=2xx((20+0)/(2))=20m`
or `s_(OABCD)=60+20=80m`
Between 0 to 2 s and 4 to 6s motion is accelerated, hence displacement-time graph is a parabola. Between 2 to 4 s motion is uniform, so displacement-time 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 figure.