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.`

Answer

A

Explanation

From `t=0` to `t=2s, a=+2m//s^2`
` :. v=at=2t`
or v-t graph is a straight line passing through origin
with slope `2 m//s^2.`
At the end of `2s,`
`v=2xx2=4 m//s`
from `t=2 to 4s, a=0.`
Hence, `v=4 m//s` will remain constant.
From `t=4` to `6s, a=-4 m/s^2.`
Hence, `v=u-at=4-4t` (with `t=0` at4s.)
`v=0` at `t=1s` or at `5s` from origin.
At the end of `6 s ( " or" t=2s) v=-4 m//s.`
Corresponding v-t graph is as shown in fig