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 ........ $s$.

$A$ three-wheeler starts from rest,accelerates uniformly with $1\; m/s^{2}$ on a straight road for $10\; s$,and then moves with uniform velocity. Plot the distance covered by the vehicle during the $n^{\text{th}}$ second $(n = 1, 2, 3, \ldots)$ versus $n$. What do you expect this plot to be during accelerated motion: a straight line or a parabola?

$A$ body moving with uniform acceleration travels a distance of $25 \ m$ in the fourth second and $37 \ m$ in the sixth second. The distance covered by the body in the next two seconds is: (in $m$)

Brakes are applied in a car moving with a speed of $20\ m/s$,which produces a retardation of $5\ m/s^2$. The distance travelled by the car until it stops is ........ $m$.

If the displacement of a particle varies with time as $\sqrt{x} = t + 7$,then

$A$ body starts from rest with uniform acceleration. If its velocity after $n$ seconds is $\upsilon$,then its displacement in the last two seconds is