$A$ body of mass $4 \,kg$ is accelerated by a constant force. It travels a distance of $5 \,m$ in the first second and a distance of $2 \,m$ in the third second. The force acting on the body is: (in $\,N$)

The velocity of a particle is $v = v_0 + gt + ft^2$. If its position is $x = 0$ at $t = 0$,then its displacement after unit time $(t = 1)$ is

$A$ particle starting from rest and moving with constant acceleration travels a distance $x$ in the first $2$ seconds and a distance $y$ in the next $2$ seconds. Then:

When will the average acceleration and instantaneous acceleration of a particle be equal over any time interval?

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)

$A$ body moves on a frictionless plane starting from rest. If $S_{n}$ is the distance moved between $t=n-1$ and $t=n$,and $S_{n-1}$ is the distance moved between $t=n-2$ and $t=n-1$,then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1-\frac{2}{x}\right)$ for $n=10$. The value of $x$ is: