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 initial velocity 4 m/s along the east and acceleration 1 $\mathrm{m}/{\mathrm{s}}^2$ along the north. The velocity of the body just after 4 s will be

1. 8 m/s along East

2. $4\sqrt{2}$ m/s along North-East

3. 8 m/s along North

4. $4\sqrt{2}$ m/s along South-East

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

1. parallel to the position vector.

2. at 60° with position vector.

3. parallel to the acceleration vector.

4. perpendicular to the position vector.

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 ${\mathrm{a}}_{\mathrm{t}}$ = kt. Distance traveled by particle on the circular path in time t is

1. $\frac{{\mathrm{kt}}^3}{3}$

2.  $\frac{{\mathrm{kt}}^2}{6}$

3.  $\frac{{\mathrm{kt}}^3}{6}$

4. $\frac{{\mathrm{kt}}^2}{2}$

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 $\mathrm{\theta }$), 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. $\frac{\mathrm{\theta }}{{\displaystyle \cos \frac{\mathrm{\theta }}{2}}}$

Two particles move from A to C and A to D on a circle of radius R and diameter AB. If the time taken by both particles are the same, then the ratio of magnitudes of their average velocities is:
                                
1. 2

2.  $2\sqrt{3}$

3.  $\sqrt{3}$

4.  $\frac{\sqrt{3}}{2}$