$A$ 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 .......... $s$ :-

$A$ particle of mass $m$ is constrained to move on the $x$-axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example,when the particle is to the left of $E$,$F$ points to the right. The magnitude of $F$ is a constant $F$ except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards the left from the equilibrium position $E$ and released from rest at $t = 0$. What is the period of the motion?

The velocity $v$ of a particle moving along the $x$-axis varies with its position $x$ as $v = \alpha \sqrt{x}$,where $\alpha$ is a constant. Which of the following graphs represents the variation of its acceleration $a$ with time $t$?

Derive the equations of motion for constant acceleration using the method of calculus.

$A$ 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)

$A$ bullet is fired into a fixed target and loses one-third of its velocity after traveling $4 \ cm$. It penetrates a further $D \times 10^{-3} \ m$ before coming to rest. The value of $D$ is: