$A$ body of mass $10 \,kg$ has a constant velocity of $10 \,m \,s^{-1}$. $A$ constant force is applied for a duration of $4 \,s$ on the object, which gives it a speed of $2 \,m \,s^{-1}$ in the opposite direction. Find the acceleration produced in it.

$A$ particle moves along $X-$axis as $x = 4(t - 2) + a(t - 2)^2$. Which of the following is true?

The acceleration $a$ (in $m/s^2$) of a body,starting from rest,varies with time $t$ (in seconds) according to the relation $a = 3t + 4$. The velocity of the body at time $t = 2 \ s$ will be $........ \ m/s$.

$A$ 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.

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:

The velocity of a bullet is reduced from $200\,m/s$ to $100\,m/s$ while travelling through a wooden block of thickness $10\,cm$. Assuming the retardation to be uniform,the retardation will be: