Two particles \(\mathrm{A}\) and \(\mathrm{B}\), move with constant velocities \(\overrightarrow{{v}_1}\) and \(\overrightarrow{{v}_2}\) respectively. At the initial moment, their position vectors are \(\overrightarrow{{r}_1}\) and \(\overrightarrow{{r}_2}\) respectively. The condition for particles \(\mathrm{A}\) and \(\mathrm{B}\) for their collision will be:
1. \(\vec{r_1} \cdot \vec{v_1}=\vec{r_2} \cdot \vec{v_2}\)
2. ​​​​​\(\dfrac{\vec{r_1}-\vec{r_2}}{\left|\vec{r_1}-\vec{r_2}\right|}=\dfrac{\vec{v_2}-\vec{v_1}}{\left|\vec{v_2}-\vec{v_1}\right|}\)${\mathrm{}}_{}$
3. \(\vec{r_1} \times \vec{v_1}=\vec{r_2} \times \vec{v_2}\)${\mathrm{ \; \; }}_{}$
4. \(\vec{r_1}-\vec{r_2}=\vec{v_1}-\vec{v_2}\) 

A point moves in a straight line so that its displacement x metre at time t sec is given by ${\mathrm{x}}^2=1+{\mathrm{t}}^2$ . Its acceleration in m/s² at time t sec is

1. $\frac{1}{{\mathrm{x}}^3}$

2. $\frac{1}{\mathrm{x}}-\frac{1}{x^2}$

3. $\frac{1}{\mathrm{x}}-\frac{t^2}{x^3}$

4. $\frac{-\mathrm{t}}{{\mathrm{x}}^2}$

A ball is falling freely strikes to ground with velocity
60 m/s. Height fallen in last one second before
hitting the ground is

1. 55 m

2. 70 m

3. 60 m

4. 80 m

The x and y coordinates of the particle at any time are $\mathrm{x}=5\mathrm{t}-2{\mathrm{t}}^2$ and y = 10 t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is

1. 0

2. 5 $\mathrm{m}/{\mathrm{s}}^2$

3. -4 $\mathrm{m}/{\mathrm{s}}^2$

4. -8 $\mathrm{m}/{\mathrm{s}}^2$

A bus moves over a straight level road with a constant acceleration a. A body in the bus drops a ball outside. The acceleration of the ball with respect to the bus and the earth are respectively

(1) a and g

(2) a + g and g - a

(3) $\sqrt{a^2+g^2} and g$

(4) $\sqrt{a^2+g^2} and a$

A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 s, the average acceleration during contact is nearly (Take g = 10 m$s^{-2}$)

(1) $1500\sqrt{2} m{s}^{-2} upwards$

(2) $1800 m{s}^{-2} downwards$

(3) $1500\sqrt{5} m{s}^{-2} upwards$

(4) $1500\sqrt{2} m{s}^{-2} downwards$

Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings (building being \(100~\text{m}\) apart and being of the same height of \(200~\text{m}\)) with the same velocity of \(25~\text{m/s}.\) When and where will the two bullets collide?
\((g = 10~\text{m/s}^2)\)

A point moves in a straight line under the retardation ${\mathrm{av}}^2$. If the initial velocity is u, the distance covered in 't' second is-

1.  aut

2.  $\frac{1}{\mathrm{a}}\ln \left(\mathrm{aut}\right)$

3.  $\frac{1}{\mathrm{a}}\ln \left(1+\mathrm{aut}\right)$

4.  $\mathrm{a} \ln \left(\mathrm{aut}\right)$

The relation between time and distance is $\mathrm{t}={\mathrm{\alpha x}}^2+\mathrm{\beta x}$, where $\mathrm{\alpha }$ and $\mathrm{\beta }$ are constants. The retardation is

1.  $2{\mathrm{\alpha v}}^3$

2.  $2{\mathrm{\beta v}}^3$

3.  $2{\mathrm{\alpha \beta v}}^3$

4.  $2{\mathrm{\beta }}^2{\mathrm{v}}^3$

A particle moves along a parabolic path \(y =9x^2\) in such a way that the \(x\) component of the velocity remains constant and has a value of \(\frac{1}{3}~\text{m/s}\)${\mathrm{}}^{}$. It can be deduced that the acceleration of the particle will be:
1. \(\frac{1}{3}\hat j~\text{m/s}^2\)
2. \(3\hat j~\text{m/s}^2\)
3. \(\frac{2}{3}\hat j~\text{m/s}^2\)
4. \(2\hat j~\text{m/s}^2\)