A man moves on a cycle with a velocity of 4 km/hr. The rain appears to fall on him with a velocity of 3 km/hr vertically. The actual velocity of the rain is

Options

A. 7 km/hr
B. 5 km/hr
C. 4/3 km/hr
D. 3/4 km/hr

Answer

To find the actual velocity of the rain, we can use vector addition. Let's break down the problem step by step.

### Step 1: Understand the velocities involved
- The man is moving on a cycle with a velocity of **4 km/hr** in the horizontal direction (let's assume this is the positive x-direction).
- The rain appears to fall vertically downwards with a velocity of **3 km/hr** relative to the man.

### Step 2: Set up the vectors
- The velocity of the man can be represented as a vector:
\[ \vec{V}_{\text{man}} = 4 \hat{i} \text{ km/hr} \]
- The velocity of the rain relative to the man is:
\[ \vec{V}_{\text{rain/man}} = -3 \hat{j} \text{ km/hr} \]
(negative because it is falling downwards).

### Step 3: Use vector addition to find the actual velocity of the rain
The actual velocity of the rain with respect to the ground can be found using the following equation:
\[ \vec{V}_{\text{rain}} = \vec{V}_{\text{rain/man}} + \vec{V}_{\text{man}} \]
Substituting the vectors we have:
\[ \vec{V}_{\text{rain}} = (-3 \hat{j}) + (4 \hat{i}) = 4 \hat{i} - 3 \hat{j} \text{ km/hr} \]

### Step 4: Calculate the magnitude of the actual velocity of the rain
To find the magnitude of the velocity vector \(\vec{V}_{\text{rain}}\), we use the Pythagorean theorem:
\[ |\vec{V}_{\text{rain}}| = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ km/hr} \]

### Conclusion
The actual velocity of the rain is **5 km/hr**.