An athlete completes one round of a circular track of radius $10 \, m$ in $40 \, sec$. The distance covered by him in $2 \, min \ 20 \, sec$ is ........ $m$.

$A$ particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{n u^2}{25 g}$,where the value of $n$ is (Given $g$ is the acceleration due to gravity).

The initial position of an object at rest is given by $3 \hat{i}-8 \hat{j}$. It moves with constant acceleration and reaches the position $2 \hat{i}+4 \hat{j}$ after $4 \, s$. What is its acceleration?

$A$ particle moves in the $XY$-plane with $x$ and $y$ coordinates varying with time $t$ as $x(t) = 5t$ and $y(t) = 5t(27 - t^2)$. At what time $t$ (in seconds) will the direction of velocity and acceleration be perpendicular to each other?

The trajectory of a particle in projectile motion is given by $y = x - \frac{x^2}{80}$. Here, $x$ and $y$ are in meters. For this projectile motion, match the following with $g = 10 \, m/s^2$.

$Column-I$	$Column-II$
$(A)$ Angle of projection	$(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$	$(q)$ $80 \, m$
$(C)$ Maximum height	$(r)$ $45^{\circ}$
$(D)$ Horizontal range	$(s)$ $\tan^{-1}(1/2)$

$A$ particle,moving with uniform speed $v$,changes its direction by angle $\theta$ in time $t$. The magnitude of its average acceleration during this time is: