The displacement $x$ (in $m$) of a particle of mass $m$ (in $kg$) moving in one dimension under the action of a force,is related to time $t$ (in $s$) by $t = \sqrt{x} + 3$. The displacement of the particle when its velocity is zero,will be

$A$ particle starts moving from rest in a straight line with constant acceleration. After time $t_0$,the acceleration changes its sign (becomes exactly opposite to the initial direction),while remaining the same in magnitude. Determine the time from the beginning of the motion at which the particle returns to its initial position.

The velocity of a bullet is reduced from $200\; m/s$ to $100\; m/s$ while travelling through a wooden block of thickness $10\; cm$. The retardation,assuming it to be uniform,will be ........... $\times 10^4\; m/s^2$.

$A$ particle starting from rest and moving with constant acceleration travels a distance $x$ in the first $2$ seconds and a distance $y$ in the next $2$ seconds. Then:

When are the acceleration and average acceleration equal for a moving object?

$A$ particle executes the motion described by $x(t) = x_0 (1 - e^{-\gamma t})$ for $t \geqslant 0$ and $x_0 > 0$.
$(a)$ Where does the particle start and with what velocity?
$(b)$ Find the maximum and minimum values of $x(t)$,$v(t)$,and $a(t)$. Show that $x(t)$ increases with time,$v(t)$ decreases with time,and $a(t)$ increases with time.