$A$ particle returns to the starting point after $10\,s$. If the rate of change of velocity during the motion is constant,then its location after $7\,s$ will be same as that after $...........\,s$.

$A$ particle moves according to the equation $y = a + bt + ct^2 - dt^4$. What are the initial velocity and initial acceleration of the particle?

Displacement $(x)$ of a particle is related to time $(t)$ as: $x = at + bt^2 - ct^3$,where $a, b$,and $c$ are constants of the motion. The velocity of the particle when its acceleration is zero is given by:

The acceleration of a particle is increasing linearly with time $t$ as $a = 6t$. The particle starts from the origin with an initial velocity $u = 10 \ m/s$. The distance travelled by the particle after $t = 2 \ s$ will be: (in $m$)

$A$ particle moves in a straight line and its position $x$ at time $t$ is given by $x^2 = 2 + t$. Its acceleration is given by

$A$ 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.