Question

The acceleration-time graph of a patticle moving in a straight line is as shown below. Sketch the v-t graph. The initial velocity of the particle is zero.

Answer 1

First find the velocities by area of the a-t graph.
From `t=0` to `t=5 s`, Area`=10xx5=50`
`v_(t=5)-v_(t=0)=50`
`impliesv_(t=5)-0=50implies v_(t=5)=50 m//s`
From `t=5` to `10 s`, Area`=-5xx5=-25`
`v_(t=10)-v_(t=5)=-25`
`implies v_(t=10)-50=-25impliesv_(t=10)=25 m//s`
Note: Area below the t-axis is negative.
From `t=0` to `t=5 s`, acceleration is constant and positive, i.e. slope of the v-t graph `(a=(dv)/(dt))` is constant and positive.
From `t=5` to `t=10 s`, acceleration is contant and negative, i.e. slope of the v-t graph is constant and negative.