The distance $x$ covered in time $t$ by a body having initial velocity $u$ and having constant acceleration $a$ is given by $x=ut+\frac{1}{2}at^2$. This result follows from

$A$ goods train accelerating uniformly on a straight railway track approaches an electric pole standing on the side of the track. Its engine passes the pole with velocity $u$ and the guard's room passes with velocity $v$. The middle wagon of the train passes the pole with a velocity of:

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

$A$ body starts from rest and moves with constant acceleration for $t$ seconds. It travels a distance $x_{1}$ in the first half of the time and $x_{2}$ in the next half of the time. Then:

The relation between time $t$ and distance $x$ of a particle is $t = ax^2 + bx$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is

What is stopping distance?