For a particle moving along a straight line path,the displacements in the third and fifth seconds of its motion are $10 \ m$ and $18 \ m$ respectively. The speed of the particle at time $t=4 \ s$ is (in $ms^{-1}$)

$A$ lift is coming down from the $8^{th}$ floor and is just about to reach the $4^{th}$ floor. Taking the ground floor as the origin and the upward direction as positive for all quantities,which one of the following is correct?

The velocity $u$ and displacement $r$ of a body are related as $u^2 = kr$,where $k$ is a constant. What will be the velocity after $1 \, s$? (Given that the displacement is zero at $t = 0$)

$A$ particle is projected with velocity $v_0$ along the $x$-axis. The deceleration of the particle is proportional to the square of the distance from the origin,i.e.,$a = -\alpha x^2$. The distance at which the particle stops is:

The velocity $v$ of a particle moving along the $x$-axis varies with its position $x$ as $v = \alpha \sqrt{x}$,where $\alpha$ is a constant. Which of the following graphs represents the variation of its acceleration $a$ with time $t$?

For a body moving with uniform acceleration along a straight line, the variation of its velocity $(v)$ with position $(x)$ is best represented by: