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.  $\sqrt{5}{\mathrm{v}}_0$

A particle starts moving on a circular path from rest, such that its tangential acceleration varies with time as \(a_t=kt\). Distance traveled by particle on the circular path in time \(t\) is:
1. \( \frac{kt^3}{3} \)
2. \(\frac{kt^2}{6} \)
3. \(\frac{kt^3}{6} \)
4. \(\frac{k t^2}{2}\)

The speed of water in a river is 4 km/h and a man can swim at 5 km/h. The minimum time taken by the man to cross the river of width 200 m is:

1.  $\frac{1}{5} \mathrm{h}$

2.  $\frac{1}{25} \mathrm{h}$

3.  $\frac{1}{15} \mathrm{h}$

4.  $\frac{1}{20} \mathrm{h}$

A particle is moving on a circular path of radius \(R.\) When the particle moves from point \(A\) to \(B\) (angle \( \theta\)), the ratio of the distance to that of the magnitude of the displacement will be:

1. $\frac{\mathrm{\theta }}{\sin {\displaystyle \frac{\mathrm{\theta }}{2}}}$

2. $\frac{\mathrm{\theta }}{2\sin {\displaystyle \frac{\mathrm{\theta }}{2}}}$

3. $\frac{\mathrm{\theta }}{2\cos {\displaystyle \frac{\mathrm{\theta }}{2}}}$

4. $\frac{\mathrm{\theta }}{{\displaystyle \cos \frac{\mathrm{\theta }}{2}}}$