An object accelerates from rest to a velocity $27.5 \,m/s$ in $10 \,s$. Find the distance covered by the object in the next $10 \,s$ (in $m$).

$A$ 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?

Particle $A$ is moving along the $x$-axis. At time $t=0$,it has a velocity of $10\,m/s$ and an acceleration of $-4\,m/s^2$. Particle $B$ has a velocity of $20\,m/s$ and an acceleration of $-2\,m/s^2$. Initially,both particles are at the origin. At time $t=2\,s$,the distance between the two particles is $.............\,m$.

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

$A$ body moves from rest with a constant acceleration of $5\,m/s^2$. Its instantaneous speed (in $m/s$) at the end of $10\,s$ is:

An object,moving with a speed of $6.25 \ m/s$,is decelerated at a rate given by: $\frac{dv}{dt} = - 2.5\sqrt{v}$,where $v$ is the instantaneous speed. The time taken by the object to come to rest would be ........ $s$.