$A$ particle moves along a straight line. Its position at any instant is given by $x = 32t - \frac{8t^3}{3}$,where $x$ is in $m$ and $t$ is in $s$. Find the acceleration of the particle at the instant when the particle is at rest $..........\,m/s^2$.

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

$A$ motorist starting a car from rest accelerates uniformly to a speed of $v \ m/s$ in $9 \ s$. He maintains this speed for another $50 \ s$ and then applies the brakes and decelerates uniformly to rest. His deceleration is numerically equal to three times his previous acceleration. Then the time during which the deceleration takes place is .......... $s$ :-

$A$ 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 $4 \,kg$ 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 $\,N$)

$A$ body falling for $2 \, s$ covers a distance $S$ equal to that covered in the next second. Taking $g = 10 \, m/s^2$,$S = .......... m$