$A$ body starts from rest with an acceleration of $8\,m/s^2$. What distance will it cover in the $5^{th}$ second?

$A$ body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20 \,m/s$ and $30 \,m/s$ respectively. The speed of the body at the midpoint of $A$ and $B$ is (nearly)

$A$ body of mass $10 \,kg$ has a constant velocity of $10 \,m \,s^{-1}$. $A$ constant force is applied for a duration of $4 \,s$ on the object, which gives it a speed of $2 \,m \,s^{-1}$ in the opposite direction. Find the acceleration produced in it.

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity ${v_0}$. The distance travelled by the particle in time $t$ will be

If the engine of a long train moving with constant acceleration crosses a tree with velocity $u$ and the last compartment of the train crosses the same tree with velocity $v$,then the velocity with which the middle compartment crosses the same tree is

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 $m \ s^{-1}$)