$A$ car moving with a velocity of $20 \,m \,s^{-1}$ is stopped in a distance of $40 \,m$. If the same car is travelling at double the velocity, the distance travelled by it for the same retardation is (in $\,m$)

$A$ car,moving with a speed of $50 \,km/hr$,can be stopped by brakes after at least $6 \,m$. If the same car is moving at a speed of $100 \,km/hr$,the minimum stopping distance is..........$m$

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:

$A$ particle starts moving along a line from zero initial velocity and comes to rest after moving a distance $d$. During its motion,it has a constant acceleration $f$ over $2/3$ of the distance and covers the rest of the distance with constant retardation. The total time taken to cover the distance 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 $t =$

If average and instantaneous velocities are same,then what is the type of motion?