If a particle moves according to the equation $v = at$,what distance will it cover in the first $4 \, s$ (in $a$)?

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?

Give an example of a motion where $x > 0$,$v < 0$,and $a > 0$ at a particular instant.

$A$ particle is projected with velocity $v_0$ along the $x-$axis,and its deceleration is given by $a = -\alpha x^2$. The distance at which the particle stops is

$A$ body starts from the origin and moves along the $X$-axis such that the velocity at any instant is given by $v = (4t^3 - 2t)$,where $t$ is in $s$ and velocity is in $m/s$. What is the acceleration of the particle when it is $2\, m$ from the origin? (in $m/s^2$)

$A$ point moves with uniform acceleration and $v_1, v_2$ and $v_3$ denote the average velocities in the three successive intervals of time $t_1, t_2$ and $t_3$. Which of the following relations is correct?