The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depend on time?

$A$ body starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the $n^{\text{th}}$ second of its motion to the total distance travelled in $n$ seconds is

$A$ 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 distance $x$ covered in time $t$ by a body having initial velocity $u$ and having constant acceleration $a$ is given by $x=ut+\frac{1}{2}at^2$. This result follows from

$A$ body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20 \,m/s$ and $30 \,m/s$ respectively. The speed of the body at the midpoint of $A$ and $B$ is (nearly)

The displacement of a particle is given by $y = a + bt + ct^2 - dt^4$. The initial velocity and acceleration are respectively