$A$ 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?

$A$ point starts moving in a straight line with a certain acceleration. At a time $t$ after the beginning of motion,the acceleration suddenly becomes a retardation of the same value. The time in which the point returns to the initial point is:

The displacement $x$ (in $m$) of a particle of mass $m$ (in $kg$) moving in one dimension under the action of a force,is related to time $t$ (in $s$) by $t = \sqrt{x} + 3$. The displacement of the particle when its velocity is zero,will be

$A$ 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)

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?

The displacement of a particle moving in a straight line is given by the expression $x = A t^3 + B t^2 + C t + D$ in metres,where $t$ is in seconds and $A, B, C$ and $D$ are constants. The ratio between the initial acceleration and initial velocity is