Question

A particle starts from rest with a time-varying acceleration $a=(2t-4)$. Here $t$ is in $s$ and $a$ in $m/s^2$. The velocity-time graph of the particle will be a

• A. parabola passing through origin.
• B. straight line not passing through origin.
• C. parabola not passing through origin
• D. straight line passing through origin

Answer 1

The correct option is A parabola passing through origin.

To find velocity, integrate acceleration w.r.t time,
$a=2t-4$
$\Rightarrow \frac{dv}{dt}=2t-4$
$\Rightarrow {\int }_0^{v}dv={\int }_0^t(2t-4)dt$
$\Rightarrow v=t^2-4t$
Hence, $v-t$ graph will be a parabola. Since initial velocity is zero, the parabola will pass through the origin.