If three coordinates of a particle change according to the equations $\mathrm{x}=3{\mathrm{t}}^2,$ $\mathrm{y}=2\mathrm{t},$ $\mathrm{z}=4$, then the magnitude of the velocity of the particle at time \(t=1\) second will be:
1. $2\sqrt{11}$ unit 
2. $\sqrt{34}$ unit
3. \(40\) unit
4. $2\sqrt{10}$ unit

Two bodies of equal masses are projected, one from top of the tower in the horizontal direction and other from the foot of tower at an angle of 45° with the horizontal, then the acceleration of their center of mass is

(1) $\frac{\mathrm{g}}{\sqrt{2}}$downward

(2) g downward

(3) $\frac{\mathrm{g}\left(\sqrt{2} - 1\right)}{\sqrt{2}}$ downward

(4) $\frac{\mathrm{g}\left(\sqrt{2} + 1\right)}{\sqrt{2}}$ upward

A projectile is projected at an angle of 60° from horizontal. At an instant, it is moving at an angle of 30° with horizontal with velocity 10 m/s. The radial acceleration of the particle at that instant is:

1.  $\frac{20}{\sqrt{2}}\mathrm{m}/{\mathrm{s}}^2$

2.  $\frac{20}{\sqrt{3}}\mathrm{m}/{\mathrm{s}}^2$

3.  $20\sqrt{3}\mathrm{m}/{\mathrm{s}}^2$

4.  $5\sqrt{3}\mathrm{m}/{\mathrm{s}}^2$

The initial velocity of a projectile is given by $\mathrm{v} = \left(20\hat{\mathrm{i}} + 50\hat{\mathrm{j}} + 70\hat{\mathrm{k}}\right)$ m/s. If acceleration due to gravity is g = 10($-\hat{\mathrm{k}}$) m/${\mathrm{s}}^2$, then the time after which projectile returns to the same horizontal level is:

(1) 4 s

(2) 10 s

(3) 14 s

(4) 12 s

Two persons start moving along two crossroads with the same speed 5 m/s. One in the direction of north and other is in the direction of the east as shown in the figure. Angular velocity of A with respect to B is :

(1) 0.5 m/s

(2) Zero

(3) 1.0 m/s

(4) 2 m/s

In projectile motion, accelerations of the projectile when it is gaining height and losing height respectively are

(1) g upward, g upward

(2) g upward, g downward

(3) g downward, g downward

(4) g downward, g upward

A person, who can swim with speed u relative to water, wants to cross a river (of width d and water is flowing with speed v). The minimum time in which the person can do so is:

1.  $\frac{\mathrm{d}}{\mathrm{v}}$

2.  $\frac{\mathrm{d}}{\mathrm{u}}$

3.  $\frac{\mathrm{d}}{\sqrt{{\mathrm{v}}^2 + {\mathrm{u}}^2}}$

4.  $\frac{\mathrm{d}}{\sqrt{{\mathrm{v}}^2 - {\mathrm{u}}^2}}$

The position vector of a particle $\overrightarrow{\mathrm{r}}$ as a function of time t (in seconds) is $\overrightarrow{\mathrm{r}}$ $=$ $\left(3\mathrm{t}\right)\hat{\mathrm{i}}$ $+$ $\left(2{\mathrm{t}}^2\right)\hat{\mathrm{j}}$ $\mathrm{m}$. The initial acceleration of the particle is:

1. 2 $\mathrm{m}/{\mathrm{s}}^2$

2. 3 $\mathrm{m}/{\mathrm{s}}^2$

3. 4 $\mathrm{m}/{\mathrm{s}}^2$

4. Zero

A particle is projected on a horizontal surface with kinetic energy K at an angle $\theta$ with the vertical. The kinetic energy of the particle at the highest point of trajectory is:

(1) K

(2) Zero

(3) K ${\cos }^2\mathrm{\theta }$

(4) K ${\sin }^2\mathrm{\theta }$

Velocity and acceleration vectors of a particle moving on a circular path are shown here. At the instant shown:

(1) The speed of the particle must be decreasing.

(2) Acceleration of particle must be increasing in magnitude.

(3) The dot product of $\overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{a}}$ is positive.

(4) The magnitude of the momentum of the particle must be increasing in magnitude.