$A$ particle moves a distance $x$ in time $t$ according to the equation $x = (t + 5)^{-1}$. The acceleration of the particle is proportional to:

$A$ body $A$ starts from rest with an acceleration $a_{1}$. After two seconds,another body $B$ starts from rest with an acceleration $a_{2}$. If they travel equal distances in the fifth second after the start of $A$,the ratio $a_{1}: a_{2}$ will be equal to:

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 $cm$)

Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.

If the velocity $v$ of a particle moving along a straight line decreases linearly with its displacement $s$ from $20 \ m/s$ to a value approaching zero at $s=30 \ m$,then the acceleration of the particle at $s=15 \ m$ is $........$

$A$ body of mass $1\,kg$ crosses a point $O$ with a velocity $60\,ms^{-1}$. $A$ force of $10\,N$ directed towards $O$ begins to act on it. It will again cross $O$ in ......... $\sec$.