$A$ particle moving along the $x-$axis has acceleration $f$ at time $t$,given by $f = f_0(1 - t/T)$,where $f_0$ and $T$ are constants. The particle at $t = 0$ has zero velocity. In the time interval between $t = 0$ and the instant when $f = 0$,the particle's velocity $(v_x)$ is:

The maximum possible acceleration of a train moving on a straight track is $10 \ m/s^2$ and maximum possible retardation is $5 \ m/s^2$. If the maximum achievable speed of the train is $10 \ m/s$,then the minimum time in which the train can complete a journey of $135 \ m$ starting from rest and ending at rest is ......... $s$.

Two cars $P$ and $Q$ are moving on a road in the same direction. The acceleration of car $P$ increases linearly with time,whereas car $Q$ moves with a constant acceleration. Both cars cross each other at time $t = 0$ for the first time. The maximum possible number of crossings (including the crossing at $t = 0$) 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

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ 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 $a$.