A particle is moving such that its position coordinates (x, y) are ($2$ m, $3$ m) at time $t=0,$ ($6$ m,$7$ m) at time $t=2$ s, and ($13$ m, $14$ m) at time $t=$ $5$ s. The average velocity vector $\vec{v}_{avg}$ from $t=$ 0 to $t=$ $5$ s is:
1. ${1 \over 5} (13 \hat{i} + 14 \hat{j})$
2. ${7 \over 3} (\hat{i} + \hat{j})$
3. $2 (\hat{i} + \hat{j})$
4. ${11 \over 5} (\hat{i} + \hat{j})$

A stone falls freely under gravity. It covers distances $h_1,~h_2$ and $h_3$ in the first $5$ seconds, the next $5$ seconds and the next $5$ seconds respectively. The relation between $h_1,~h_2$ and $h_3$ is:

A particle has initial velocity $\left(2 \hat{i} + 3 \hat{j}\right)$ and acceleration $\left(0 . 3 \hat{i} + 0 . 2 \hat{j}\right)$. The magnitude of velocity after $10$ s will be:

The motion of a particle along a straight line is described by the equation $x = 8+12t-t^3$ where $x$ is in meter and $t$ in seconds. The retardation of the particle, when its velocity becomes zero, is:
1. $24$ ms^-2
2. zero
3. $6$ ms^-2
4. $12$ ms^-2

A particle covers half of its total distance with speed ν₁ and the rest half distance with speed ν₂.
Its average speed during the complete journey is:

1. $\frac{v_1+v_2}{2}$

2. $\frac{v_1{v}_2}{v_1+v_2}$

3. $\frac{2v_1{v}_2}{v_1+v_2}$

4. $\frac{v_1^2{v}_2^2}{v_1^2+v_2^2}$

A ball is dropped from a high-rise platform at $t=0$ starting from rest. After $6$ seconds, another ball is thrown downwards from the same platform with speed $v$. The two balls meet after $18$ seconds. What is the value of $v$?

A particle moves a distance $x$ in time $t$ according to equation $x=(t+5)^{-1}.$ The acceleration of the particle is proportional to:
1. (velocity)^$3/2$
2. (distance)^$2$
3. (distance)^$-2$
4. (velocity)^$2/3$