The acceleration versus time graph of a particle moving along a straight line is shown in the figure. Draw the respective velocity-time graph Given `v=0` at `t=0.`

Explanation

From `t = 0 to t = 2, a = + 2 m//s^(2) rArr v = at = 2t`
or v-t graph is a straight line passing through origin with slope `2 m//s^(2)`.
At the end of 2 s,
`v=2xx2=4 m//s`
From t = 2 to 4 s, a = 0.
Hence, v = 4 m/s will remain constant.
From `t = 4 6 s, a=-4 m//s^(2)`. Hence,
`v=u-at=4-4t` (with t =0 at 4 s)
v = 0 at t = 1 or at 5 s from origin.
At the end of 6 s (or t = 2 s) v = - 4 m/s. Corresponding v-t graph is sa shown below