The acceleration of a particle is increasing linearly with time $t$ as $a = 6t$. The particle starts from the origin with an initial velocity $u = 10 \ m/s$. The distance travelled by the particle after $t = 2 \ s$ will be: (in $m$)

The velocity of a bullet is reduced from $200\,m/s$ to $100\,m/s$ while travelling through a wooden block of thickness $10\,cm$. Assuming the retardation to be uniform,the retardation will be:

Two bodies $A$ and $B$ move in the same straight line starting from the same position. Body $A$ moves with a constant velocity $u$ and body $B$ moves with a constant acceleration $a$ starting from rest. When their velocities become equal,the distance between them is:

The velocity of a body moving with a uniform acceleration of $2\,m/s^2$ is $10\,m/s$. Its velocity after an interval of $4\,s$ is...........$m/s$.

$A$ particle is moving with speed $v = b\sqrt{x}$ along the positive $x$-axis. Calculate the speed of the particle at time $t = \tau$ (assume that the particle is at the origin at $t = 0$).

The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $\omega = a \sqrt{v}$, where $a$ is a constant. If the initial velocity of the car is $v_0$, the distance the car will travel and the time it takes before it stops are: