What is the relation between displacement,time,and acceleration for a body undergoing uniform acceleration?

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 $a$.

Two bodies $A$ and $B$ start from the same point at the same instant and move along a straight line. Body $A$ moves with uniform acceleration $a$ and body $B$ moves with uniform velocity $V$. They meet after time $t$. The value of $t$ is

The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $\omega = a \sqrt{v}$, where $a$ is a constant. If the initial velocity of the car is $v_0$, the distance the car will travel and the time it takes before it stops are:

$A$ body travels $102.5 \ m$ in the $n^{\text{th}}$ second and $115.0 \ m$ in the $(n+2)^{\text{th}}$ second. The acceleration is: (in $m/s^2$)

For a train engine moving with a speed of $20 \; m/s$,the driver must apply brakes at a distance of $500 \; m$ before the station for the train to come to rest at the station. If the brakes were applied at half of this distance,the train engine would cross the station with a speed of $\sqrt{x} \; m/s$. The value of $x$ is $..............$ (Assuming the same retardation is produced by the brakes).