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Velocity-time graph of a particle moving...

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.`

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The correct Answer is:
A

Displacement =Area under velocity-time graph
Hence, `s_(OA)=1/2xx2xx10=10m`
`s_(AB)=2xx10=20m`
or `s_(OAB)=10+20=30m`
`s_(BC)=1/2xx2(10+20)=30m`
or `s_(OABC)=30+30=60m`
and `s_(CD)=1/2xx2xx20=20m`
or `s_(OABCD)=60+20=80m`

Between 0 to 2 s an 4 to 6s motion is accelerated.
hence displacement-time graph is a parabola.
Between 2 to 4s motion is uniform, so
displacement-time graph will be a straight line.
Between 6 to 8s motion is declerated hence
displacement-time graph is given 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 above
figure.
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