The velocity $(v)$ of a particle starting from rest increases linearly with time $(t)$ as $v = 4t$,where $v$ is in $m s^{-1}$ and $t$ is in seconds. The distance covered by the particle in the first $4$ seconds is (in $m$)

$A$ body starts moving from rest with constant acceleration. It covers a displacement $S_1$ in the first $(p-1)$ seconds and $S_2$ in the first $p$ seconds. The total displacement $S_1+S_2$ will be covered in time:

$A$ 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:

$A$ body of mass $10 \, kg$ is moving with a constant velocity of $10 \, m/s$. When a constant force acts for $4 \, s$ on it,it moves with a velocity of $2 \, m/s$ in the opposite direction. The acceleration produced in it is .......... $m/s^2$.

What kind of motion is represented by the equation $\Delta S = v \Delta t$?

$A$ vehicle starts moving in a straight line with an acceleration $a = 4 \,m/s^2$, with initial velocity equal to zero. After accelerating for time $t_1$, the vehicle moves uniformly for time $t_2$, and finally decelerates for time $t_1$, eventually coming to a stop. The total time taken during the motion is $10 \,s$ and the average velocity during the motion is $5.1 \,m/s$. The time taken by the vehicle during acceleration is (in $\,s$)