Question

A person walks at the rate of 3 km/hr. Rain appears to him in vertical direction at the rate of $3\sqrt{3}$ km/hr. Find magnitude and direction of true velocity of rain.

• A. 6 km/hr, inclined at an angle of $45^\circ$ to the vertical towards the person's motion.
• B. 3 km/hr, inclined at an angle of $30^\circ$ to the vertical towards the person's motion.
• C. 6 km/hr, inclined at an angle of $30^\circ$ to the vertical towards the person's motion.
• D. 6 km/hr, inclined at an angle of $60^\circ$ to the vertical towards the person's motion.

Answer 1

Answer: (C)

6 km/hr, inclined at an angle of $30^\circ$ to the vertical towards the person's motion.

Answer 2

To solve this problem, we use the concept of relative velocity.

Let:

• $\vec{v}_p$ be the velocity of the person.
• $\vec{v}_r$ be the true velocity of the rain.
• $\vec{v}_{r/p}$ be the apparent velocity of the rain with respect to the person.

The relationship between these velocities is given by the relative velocity formula: $\vec{v}_{r/p} = \vec{v}_r - \vec{v}_p$ We need to find the true velocity of the rain, $\vec{v}_r$. Rearranging the formula, we get: $\vec{v}_r = \vec{v}_{r/p} + \vec{v}_p$

Let's define a coordinate system:

• Let the positive x-axis be the direction of the person's motion (horizontal).
• Let the positive y-axis be vertically upwards, so the negative y-axis is vertically downwards.

Given information:

1. Velocity of the person ($\vec{v}_p$): The person walks at 3 km/hr. $\vec{v}_p = 3 \hat{i} \text{ km/hr}$
2. Apparent velocity of rain ($\vec{v}_{r/p}$): Rain appears to him in the vertical direction at the rate of $3\sqrt{3}$ km/hr. Since rain falls downwards, the apparent velocity is in the negative y-direction. $\vec{v}_{r/p} = -3\sqrt{3} \hat{j} \text{ km/hr}$

Now, substitute these vector components into the equation for $\vec{v}_r$: $\vec{v}_r = (-3\sqrt{3} \hat{j}) + (3 \hat{i})$ $\vec{v}_r = 3 \hat{i} - 3\sqrt{3} \hat{j} \text{ km/hr}$

Magnitude of the true velocity of rain: The magnitude of $\vec{v}_r$ is given by: $|\vec{v}_r| = \sqrt{(v_{rx})^2 + (v_{ry})^2}$ $|\vec{v}_r| = \sqrt{(3)^2 + (-3\sqrt{3})^2}$ $|\vec{v}_r| = \sqrt{9 + (9 \times 3)}$ $|\vec{v}_r| = \sqrt{9 + 27}$ $|\vec{v}_r| = \sqrt{36}$ $|\vec{v}_r| = 6 \text{ km/hr}$

Direction of the true velocity of rain: The true velocity vector is $\vec{v}_r = 3 \hat{i} - 3\sqrt{3} \hat{j}$. This means the rain has a horizontal component of 3 km/hr in the direction of the person's motion and a vertical component of $3\sqrt{3}$ km/hr downwards.

Let $\theta$ be the angle that $\vec{v}_r$ makes with the vertical direction (negative y-axis). We can use the tangent function: $\tan \theta = \frac{\text{Horizontal component}}{\text{Vertical component magnitude}}$ $\tan \theta = \frac{|v_{rx}|}{|v_{ry}|} = \frac{3}{3\sqrt{3}}$ $\tan \theta = \frac{1}{\sqrt{3}}$ Therefore, $\theta = 30^\circ$ Since the horizontal component ($3 \hat{i}$) is in the same direction as the person's motion, the rain is inclined at an angle of $30^\circ$ to the vertical towards the person's motion.

Thus, the true velocity of rain is 6 km/hr, inclined at an angle of $30^\circ$ to the vertical towards the person's motion.