If the velocity of a particle moving along a straight line with uniform acceleration is given by $V = \sqrt{196 - 16x} \text{ m/s}$, then its acceleration is ($x$ is the displacement of the particle). (in $\text{ m/s}^2$)

Two particles $A$ and $B$ start from rest and move for equal time on a straight line. Particle $A$ has an acceleration of $2\,m/s^2$ for the first half of the total time and $4\,m/s^2$ for the second half. The particle $B$ has acceleration $4\,m/s^2$ for the first half and $2\,m/s^2$ for the second half. Which particle has covered a larger distance?

If a body starts from rest and travels $120 \, cm$ in the $6^{th}$ second,then the acceleration of the body 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:

The velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$,then at $t = 2 \ s$,its position is: (in $m$)

$A$ truck moving with a constant velocity $12 \,m/s$ crosses a car moving from rest with uniform acceleration $2 \,m/s^2$. The distance the car has to travel from the starting point to cross the truck again is (in $\,m$)