The acceleration $a$ in $m/s^2$ of a particle is given by $a = 3t^2 + 2t + 2$,where $t$ is the time. If the particle starts with an initial velocity $u = 2\,m/s$ at $t = 0$,then the velocity at the end of $2\,s$ is:

$A$ body $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ starts from the same point and moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is:

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)

If a body starts from rest and travels $120 \, cm$ in the $6^{th}$ second,then the acceleration of the body is:

$A$ car accelerates from rest at a constant rate $\alpha$ for some time,after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$,then the maximum velocity acquired by the car is

Object $A$ starts from rest with a constant acceleration $a$. Object $B$ starts from the same position and moves in the same direction as $A$ with a constant velocity $v$. If both meet after time $t$,then $t =$