Two stones are thrown up vertically and simultaneously but with different speeds. Which graph correctly represents the time variation of their relative positions $\Delta x$.Assume that stones do not bounce after hitting ground.

A particle starts from rest and performing circular motion of constant radius with speed given by $v = \alpha \sqrt x$ where $\alpha$ is a constant and $x$ is the distance covered. The correct graph of magnitude of its tangential acceleration $(a_t)$ and centripetal acceleration $(a_c)$ versus $t$ will be:

If a particle takes $t$ second less and acquires a velocity of $v \ ms^{^{-1}}$ more in falling through the same distance (starting from rest) on two planets where the accelerations due to gravity are $2 \,\, g$ and $8 \,\,g$ respectively then $v=$

A rigid rod is sliding. At some instant position of the rod is as shown in the figure. End $A$ has constant velocity $v_0$. At $t = 0, y = l$ . 

At time $t =0$ a particle starts travelling from a height $7\,\hat{z} cm$ in a plane keeping $z$ coordinate constant. At any instant of time it's position along the $x$ and $y$ directions are defined as $3\,t$ and $5\,t^{3}$ respectively. At $t =1\,s$ acceleration of the particle will be.

For any arbitrary motion in space, which of the following relations are true

$(a)$ $\left. v _{\text {average }}=(1 / 2) \text { (v }\left(t_{1}\right)+ v \left(t_{2}\right)\right)$

$(b)$ $v _{\text {average }}=\left[ r \left(t_{2}\right)- r \left(t_{1}\right)\right] /\left(t_{2}-t_{1}\right)$

$(c)$ $v (t)= v (0)+ a t$

$(d)$ $r (t)= r (0)+ v (0) t+(1 / 2)$ a $t^{2}$

$(e)$ $a _{\text {merage }}=\left[ v \left(t_{2}\right)- v \left(t_{1}\right)\right] /\left(t_{2}-t_{1}\right)$

(The 'average' stands for average of the quantity over the time interval $t_{1}$ to $t_{2}$ )