Velocity of a body moving along a straight line with uniform acceleration $a$ reduces by $\frac{3}{4}$ of its initial velocity in time $t_0$. The total time of motion of the body till its velocity becomes zero is

$A$ 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:

$A$ 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:

$A$ car accelerates from rest at a constant rate $\alpha$ for some time,after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$,then the maximum velocity acquired by the car is

Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.

The acceleration of an object increases with time as $a = bt$. The object starts from the origin with an initial velocity $v_0$. Find the distance traveled by the object in time $t$.