A particle is moving in a straight line and its acceleration-time graph is shown below. At `t=0`, the particle is at rest. Find the velocity of the particle at `t=3 s, 6 s` and `8 s`. Also sketch the v-s graph.

Explanation

Given `t=0, v_(t=0)=0`

Area for the a-t graph for `t =0` to `t=6 s=1/2xx10xx6=30`
`v_(t=6)-v_(t=0)=30 implies v_(t=6)=30 m//s`

From `t=0` to `t=8 s`, Area`=1/2xx10xx6+10xx2=50`
`v_(t=8)-v_(t=0)=50impliesv_(t=8)=50 m//s`
To calculate the velocity at `t=3 s`, first find the acceleration at `t=3 s`

`tan alpha=a_(1)/3=10/6 implies a_(1)=5m//s^(2)`

from `t=0` to `t=3 s`, Area`=1/2xxa_(1)xx3=1/2xx5xx3=7.5`
`v_(t=3)-v_(t=0)=7.5implies v_(t=3)=7.5 m//s`
`v_(t=10)-v_(t=0)=`Area for `t=0` to `t=10 s=70`
`v_(t=10)=70 m//s`
From `t=0` to `t=6 s`, a is +ve and increase and hence, the slope of the v-t graph is +ve and increasing.
From `t=6` to `t=10 s`, a is +ve and constant and hence, the slope of the v-t graph is +ve and constant, graph will be an inclined straight line.