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

$A$ body starts from rest with uniform acceleration $a$. Its velocity after $n \ s$ is $v$. The displacement of the body in the last $3 \ s$ is: (Assume total time of journey is $n \ s$)

$A$ particle initially at rest is moving along a straight line with an acceleration of $2 \,m/s^2$. At a time of $3 \,s$ after the beginning of motion, the direction of acceleration is reversed. The time from the beginning of the motion in which the particle returns to its initial position is

Two cars $A$ and $B$ are initially at rest at the same point. If car $A$ starts with a uniform velocity of $40\, m/s$ and car $B$ starts in the same direction with a constant acceleration of $4\, m/s^2$,then after how much time will car $B$ catch car $A$?

Consider a vehicle moving with a velocity $54 \,km/h$. At a distance of $400 \,m$ from the traffic light, brakes are applied. The acceleration of the vehicle after the application of brakes is $-0.3 \,m/s^2$. The vehicle's position relative to the traffic light is: (in $\,m$)

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 $t =$