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

In a uniform circular motion, if the speed of the particle is 2 m/s and radius of the circle is 2 m, then the values of centripetal and tangential acceleration are, respectively:

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

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

3. 0, 2 $\mathrm{m}/{\mathrm{s}}^2$

4. 2 $\mathrm{m}/{\mathrm{s}}^2$, 0

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

Coordinates of a particle as a function of time t are x = 2t, y = 4t. It can be inferred that the path of the particle will be:

1. Straight line

2. Ellipse

3. Parabola

4. Hyperbola

When a man walks on a horizontal road with velocity 1 km/h, the rain appears to him coming vertically at a speed of 2 km/h. The actual speed of the rain with respect to ground is:

1. $\sqrt{3}$ km/h

2. $\sqrt{5}$ km/h

3. 1 km/h

4. 3 km/h

A body started moving with an initial velocity of \(4\) m/s along the east and an acceleration \(1\) m/s² along the north. The velocity of the body just after \(4\) s will be?

A particle is thrown obliquely at \(t=0\). The particle has the same K.E. at \(t=5\) seconds and at \(t=9\) seconds. The particle attains maximum altitude at:
1. \(t=6\) s
2. \(t=7\) s
3. \(t=8\) s
4. \(t=14\) s

The position vector of a particle is $\overrightarrow{\mathrm{r}}$ $=$ $\left(\mathrm{a}\right)$ $\sin$ $\mathrm{\omega t}\hat{\mathrm{i}}$ $+$ $\left(\mathrm{a}\right)$ $\cos$ $\mathrm{\omega t}\hat{\mathrm{j}}$. The velocity of the particle is:

A projectile is projected from the ground with the velocity ${\mathrm{v}}_0$ at an angle $\theta$ with the horizontal. What is the vertical component of the velocity of the projectile when its vertical displacement is equal to half of the maximum height attained?

1.  $\sqrt{3}{\mathrm{v}}_0$ $\mathrm{cos\theta }$

2.  $\frac{{\mathrm{v}}_0}{\sqrt{2}}\sin$ $\mathrm{\theta }$

3.  $\frac{{\mathrm{v}}_0}{\sqrt{2}}\cos \theta$

4.  $\sqrt{5}{\mathrm{v}}_0$