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:

Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.

$A$ body $A$ starts from rest with an acceleration $a_{1}$. After two seconds,another body $B$ starts from rest with an acceleration $a_{2}$. If they travel equal distances in the fifth second after the start of $A$,the ratio $a_{1}: a_{2}$ will be equal to:

$A$ body of mass $10 \, kg$ is moving with a constant velocity of $10 \, m/s$. When a constant force acts for $4 \, s$ on it,it moves with a velocity of $2 \, m/s$ in the opposite direction. The acceleration produced in it is .......... $m/s^2$.

$A$ particle moves in a straight line with a constant acceleration. It changes its velocity from $10 \, m/s$ to $20 \, m/s$ while passing through a distance of $135 \, m$ in $t$ seconds. The value of $t$ is..........$s$.

$A$ lift is coming down from the $8^{th}$ floor and is just about to reach the $4^{th}$ floor. Taking the ground floor as the origin and the upward direction as positive for all quantities,which one of the following is correct?