Consider the motion of the tip of the second hand of a clock. In one minute (assuming \(R\) to be the length of the second hand), its:

1.	displacement is \(2\pi R\)
2.	distance covered is \(2R\)
3.	displacement is zero.
4.	distance covered is zero.

A person, reaches a point directly opposite on the other bank of a flowing river, while swimming at a speed of 5 m/s at an angle of \(120^\circ\) with the flow. The speed of the flow must be:
1. \(2.5\) m/s
2. \(3\) m/s
3. \(4\) m/s
4. \(1.5\) m/s

A car with a vertical windshield moves in a rain storm at a speed of \(40\) km/hr. The rain drops fall vertically with a constant speed of \(20\) m/s. The angle at which raindrops strike the windshield is:
1. ${\tan }^{-1}\frac{5}{9}$

2. ${\tan }^{-1}\frac{9}{5}$

3. ${\tan }^{-1}\frac{3}{2}$

4. ${\tan }^{-1}\frac{2}{3}$

A particle projected from origin moves in the x-y plane with a velocity $\overrightarrow{\mathrm{v}}=3\hat{\mathrm{i}}+6\mathrm{x}\hat{\mathrm{j}}$, where $\hat{\mathrm{i}}$ and $\hat{\mathrm{j}}$ are the unit vectors along the x and y-axis. The equation of path followed by the particle is:

1. $\mathrm{y}={\mathrm{x}}^2$

2. $\mathrm{y}=\frac{1}{{\mathrm{x}}^2}$

3. $\mathrm{y}=2{\mathrm{x}}^2$

4. $\mathrm{y}=\frac{1}{\mathrm{x}}$

The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by $\mathrm{y}=(4\mathrm{t}-2{\mathrm{t}}^2)\mathrm{m}$ and $\mathrm{x}=\left(3\mathrm{t}\right)$ metre, where t is in seconds and point of projection is taken as the origin. The angle of projection of projectile with vertical is:

1. $30°$

2. $37°$

3. $45°$

4. $60°$

The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection (u). Its range on the horizontal plane is:

1. $\frac{\sqrt{3}{\mathrm{u}}^2}{2\mathrm{g}}$

2. $\frac{3{\mathrm{u}}^2}{2\mathrm{g}}$

3. $\frac{3{\mathrm{u}}^2}{\mathrm{g}}$

4. $\frac{{\mathrm{u}}^2}{2\mathrm{g}}$

The equation of a projectile is $\mathrm{y}=\mathrm{ax}-{\mathrm{bx}}^2$. Its horizontal range is?

1. $\frac{\mathrm{a}}{\mathrm{b}}$

2. $\frac{\mathrm{b}}{\mathrm{a}}$

3. $\mathrm{a}+\mathrm{b}$

4. $\mathrm{b}-\mathrm{a}$

When a particle is projected at some angle to the horizontal, it has a range R and time of flight t₁. If the same particle is projected with the same speed at some other angle to have the same range, its time of flight is t₂, then:

1. ${\mathrm{t}}_1+{\mathrm{t}}_2=\frac{2\mathrm{R}}{\mathrm{g}}$

2. ${\mathrm{t}}_1-{\mathrm{t}}_2=\frac{\mathrm{R}}{\mathrm{g}}$

3. ${\mathrm{t}}_1{\mathrm{t}}_2=\frac{2\mathrm{R}}{\mathrm{g}}$

4. ${\mathrm{t}}_1{\mathrm{t}}_2=\frac{\mathrm{R}}{\mathrm{g}}$

A car is moving at a speed of 40 m/s on a circular track of radius 400 m. This speed is increasing at the rate of 3 m/s². The acceleration of the car is:

1. 4 m/s²

2. 7 m/s²

3. 5 m/s²

4. 3 m/s²

A particle is moving along a circle such that it completes one revolution in 40 seconds. In 2 minutes 20 seconds, the ratio of \(|displacement| \over distance\) will be:
1. 0
2. 1/7
3. 2/7
4. 1/11