$A$ body starts from rest and moves with constant acceleration for $t$ seconds. It travels a distance $x_{1}$ in the first half of the time and $x_{2}$ in the next half of the time. Then:

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 $a$.

$A$ particle starts with an initial velocity of $10.0 \, m/s$ along the $x$-direction and accelerates uniformly at the rate of $2.0 \, m/s^2$. The time taken by the particle to reach the velocity of $60.0 \, m/s$ is $....... \, s$.

The position of a particle moving along the $x$-axis is given by $x = (-2t^3 + 3t^2 + 5) \ m$. The acceleration of the particle at the instant its velocity becomes zero is ....... $m/s^2$.

$A$ 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:

$A$ man of height $h$ is walking away from a street lamp with a constant speed $v$. The height of the street lamp is $3h$. The rate at which the length of the man's shadow is increasing when he is at a distance $10h$ from the base of the street lamp is: