Question

To a man walking at the rate of 3 km/h, the rain appears to fall vertically. When he increases his speed to 6 km/h, it appears to meet him at an angle of 45° with vertical. The speed of rain is

• A. 2$\sqrt{3}$ km/h
• B. $\sqrt{6}$ km/h
• C. 3$\sqrt{2}$ km/h
• D. 4$\sqrt{3}$ km/h

Answer 1

Answer: (C)

3$\sqrt{2}$ km/h

Answer 2

The problem involves relative motion, specifically the velocity of rain relative to a moving man. We can use vector addition/subtraction to solve this.

Let $\vec{v}_R$ be the actual velocity of rain and $\vec{v}_M$ be the velocity of the man. The velocity of rain relative to the man is given by $\vec{v}_{R/M} = \vec{v}_R - \vec{v}_M$.

Let's assume the man walks along the positive x-axis and the rain has a horizontal component $v_x$ and a vertical component $v_y$. So, $\vec{v}_R = v_x \hat{i} + v_y \hat{j}$. Since rain falls downwards, $v_y$ will be negative.

Scenario 1:

The man walks at $3 \text{ km/h}$. So, $\vec{v}_M = 3 \hat{i}$.

The rain appears to fall vertically, which means its relative horizontal component is zero.

The relative velocity is $\vec{v}_{R/M} = (v_x \hat{i} + v_y \hat{j}) - 3 \hat{i} = (v_x - 3) \hat{i} + v_y \hat{j}$.

Since the rain appears to fall vertically, the horizontal component $(v_x - 3)$ must be zero.

$v_x - 3 = 0 \Rightarrow v_x = 3 \text{ km/h}$.

So, the actual velocity of rain is $\vec{v}_R = 3 \hat{i} + v_y \hat{j}$.

Scenario 2:

The man increases his speed to $6 \text{ km/h}$. So, $\vec{v}_M' = 6 \hat{i}$.

The rain appears to meet him at an angle of $45^\circ$ with the vertical.

The new relative velocity is $\vec{v}_{R/M}' = \vec{v}_R - \vec{v}_M' = (3 \hat{i} + v_y \hat{j}) - 6 \hat{i} = (3 - 6) \hat{i} + v_y \hat{j} = -3 \hat{i} + v_y \hat{j}$.

Let the horizontal component of $\vec{v}_{R/M}'$ be $v_{R/M,x}' = -3 \text{ km/h}$ and the vertical component be $v_{R/M,y}' = v_y$.

The angle $\theta$ with the vertical is given by $\tan \theta = \frac{|v_{R/M,x}'|}{|v_{R/M,y}'|}$.

Given $\theta = 45^\circ$, we have $\tan 45^\circ = \frac{|-3|}{|v_y|}$.

$1 = \frac{3}{|v_y|} \Rightarrow |v_y| = 3 \text{ km/h}$.

Since rain falls downwards, $v_y = -3 \text{ km/h}$.

Now we have both components of the actual velocity of rain:

$\vec{v}_R = 3 \hat{i} - 3 \hat{j}$.

The speed of rain is the magnitude of $\vec{v}_R$:

Speed of rain $= |\vec{v}_R| = \sqrt{v_x^2 + v_y^2} = \sqrt{(3)^2 + (-3)^2}$

Speed of rain $= \sqrt{9 + 9} = \sqrt{18}$

Speed of rain $= \sqrt{9 \times 2} = 3\sqrt{2} \text{ km/h}$.