The displacement of a particle after time $t$ is given by $x = (k/b^2)(1 - e^{-bt})$ where $b$ is a constant. What is the acceleration of the particle?

$A$ 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.

An object moving along the $x$-axis with a uniform acceleration has a velocity $\vec{v} = (12 \ cm \ s^{-1}) \hat{i}$ at $x = 3 \ cm$. After $2 \ s$,if it is at $x = -5 \ cm$,then its acceleration is:

$A$ particle starts from origin $O$ from rest and moves with a uniform acceleration along the positive $x$-axis. Identify all figures that correctly represent the motion qualitatively. ($a =$ acceleration,$v =$ velocity,$x =$ displacement,$t =$ time)

$A$ particle is moving along a straight line with constant acceleration. At the end of the $10^{th}$ second,its velocity becomes $20 \, m/s$,and in the $10^{th}$ second,it travels a distance of $10 \, m$. Then the acceleration of the particle will be ........ $m/s^2$.

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: