The relation between time $t$ and distance $x$ for a moving body is given as $t = m x^{2} + n x$,where $m$ and $n$ are constants. The retardation of the motion is:

If the displacement of a particle varies with time as $\sqrt{x} = t + 7$,then

$A$ body starts from rest. What is the ratio of the distance travelled by the body during the $4^{th}$ and $3^{rd}$ second?

$A$ body starts from rest with uniform acceleration. If its velocity after $n^{\text{th}}$ second (last second) is $V$, then its displacement in the last two seconds 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:

$A$ 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: