Tripling the speed of the motor car multiplies the distance needed for stopping it by

$A$ car moving along a straight highway with a speed of $126 \; km/h$ is brought to a stop within a distance of $200 \; m$. What is the retardation of the car (assumed uniform in $m/s^2$)?

$A$ truck moving with a constant velocity $12 \,m/s$ crosses a car moving from rest with uniform acceleration $2 \,m/s^2$. The distance the car has to travel from the starting point to cross the truck again is (in $\,m$)

The velocity $u$ and displacement $r$ of a body are related as $u^2 = kr$,where $k$ is a constant. What will be the velocity after $1 \, s$? (Given that the displacement is zero at $t = 0$)

$A$ body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20 \,m/s$ and $30 \,m/s$ respectively. The speed of the body at the midpoint of $A$ and $B$ is (nearly)

$A$ 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 $\text{ cm}$)