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

The relation between the position $x$ and time $t$ of a particle moving in one dimension is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. When the velocity of the particle becomes zero,the displacement of the particle is ........ $m$.

$A$ particle moves along a straight line $OX$. At a time $t$ (in seconds),the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?

$A$ moving body is covering distances which are proportional to the square of the time. Then the acceleration of the body is

$A$ 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)

$A$ body initially at rest is moving with uniform acceleration $a$. Its velocity after $n$ seconds is $v$. The displacement of the body in the last $2 \ s$ is