$A$ body starts from rest and acquires a velocity of $10 \ m \ s^{-1}$ in $2 \ s$. What is the acceleration of the body and the distance travelled?

The same retarding force is applied to stop a train. The train stops after $80 \ m$. If the speed is doubled,then the stopping distance will be

$A$ body starts with an initial velocity of $10\,m/s$ and moves along a straight line with constant acceleration. When the velocity of the particle is $50\,m/s$,the acceleration is reversed in direction. The velocity of the particle when it again reaches the starting point is $............\,m/s$.

$A$ body starting from rest moves with an acceleration of $\frac{5}{4} \,ms^{-2}$. The distance travelled by the body in the third second is:

$A$ body travels for $15\, s$ starting from rest with constant acceleration. If it travels distances ${S_1}$,${S_2}$,and ${S_3}$ in the first $5\, s$,the second $5\, s$,and the next $5\, s$ respectively,the relation between ${S_1}$,${S_2}$,and ${S_3}$ is:

$A$ 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 $m$)