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

$A$ particle experiences a constant acceleration for $20\, s$ after starting from rest. If it travels a distance $s_1$ in the first $10\, s$ and a distance $s_2$ in the next $10\, s$,then:

Two cars $A$ and $B$ are initially at rest at the same point. If car $A$ starts with a uniform velocity of $40\, m/s$ and car $B$ starts in the same direction with a constant acceleration of $4\, m/s^2$,then after how much time will car $B$ catch car $A$?

$A$ particle starts its motion from rest under the action of a constant force. If the distance covered in the first $10 \ s$ is $S_1$ and that covered in the first $20 \ s$ is $S_2$,then:

$A$ particle starts moving from rest in a straight line with constant acceleration. After time $t_0$,the acceleration changes its sign (becomes exactly opposite to the initial direction),while remaining the same in magnitude. Determine the time from the beginning of the motion at which the particle returns to its initial position.

If a body loses half of its velocity on penetrating $3 \, cm$ in a wooden block,then how much will it penetrate more before coming to rest? (in $cm$)