Question

A particle is moving rectilinearly so that its acceleration is given as $a\left(\text{in}m/s^2\right)=3t^2+1$. Its initial velocity is zero then

• A. The displacement of the particle in $1s$ will be $2m$
• B. The velocity of the particle at $t=1s$ will be $2m/s$
• C. The particle will continue to move in positive direction
• D. The particle will come back to its starting point after some time.

Answer 1

Answer: (B), (C)

The correct options are
B The velocity of the particle at $t=1s$ will be $2m/s$
C The particle will continue to move in positive direction

$a=3t^2+1$
$\Rightarrow \frac{dv}{dt}=3t^2+1$
$\Rightarrow \underset{0}{\overset{v}{\int }}dv=\underset{0}{\overset{t}{\int }}(3t^2+1)dt$
$v=(t^3+t{)}_0^t$
$\Rightarrow v=t^3+t$
At $t=1s$,
$v=1+1=2m/s$
$v=t^3+t$
$\frac{ds}{dt}=t^3+t$
$\Rightarrow \underset{0}{\overset{S}{\int }}ds=\underset{0}{\overset{1}{\int }}(t^3+t)dt={[\frac{t^4}{4}+\frac{t^2}{2}]}_0^1$
$\Rightarrow S=\frac{1}{4}+\frac{1}{2}=0.75m$
As both $v$ and $a$ will continue to increase with increasing $t$, thus it will continue to move in the positive direction and will come never come back to its starting point.