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)xx2xx10=10 m`
`s_(AB)=2xx10=20 m`
or `s_(OAB)=10+20=30 m`
`s_(BC)=(1)/(2)xx2(10+20)=30 m`
or `s_(OABC)=30+30=60 m`
and `s_(CD)=(1)/(2)xx20=20 m`
or `s_(OABCD)=60+20=80 m`

Between 0 to 2 and 4 to 6 s motion is accelerated, hence displacement - time graph is a parabola. Between 2 to 4 s motion is uniform, so displacement - time graph will be a straight l,ine. Between 6 to 8 motion is decelerated hence displacement time graph is gain a parabola but inverted in shape. At the of 8 s velocity is zero, therefore, slope of displacement - time graph should be zero. The corresponding graph is shown in figure.