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 is moving from rest under constant acceleration. Let $S_1$ be the displacement in the first $(p - 1) \ s$ and $S_2$ be the displacement in the first $p \ s$. The displacement in the $(p^2 - p + 1)^{th} \ s$ will be:

If a car at rest accelerates uniformly to a speed of $144 \, km/h$ in $20 \, s$,then it covers a distance of ........ $m$.

Two cars $A$ and $B$ initially at rest are moving in the same direction with accelerations $a_1$ and $a_2$ respectively. After a certain time $t$,they achieve velocities $v_1$ and $v_2$ respectively and are separated by a distance of $50 \ m$. If $(a_1 - a_2) = 4 \ m \ s^{-2}$,then the quantity $(v_1 - v_2)$ will be: (in $m \ s^{-1}$)

The displacement of a particle starting from rest at $t=0$ is given by $s=9 t^2-2 t^3$. The time in seconds at which the particle will attain zero velocity is (in $s$)

$A$ particle with initial velocity $v_0$ moves with constant acceleration $a$ in a straight line. Find the distance travelled in the $n^{th}$ second.