Mohan and Sohan walk at average speed of 1 km/hr and 1.5 km/hr respectively. If Mohan walks in South direction and Sohan in East direction beginning from the same origin at the same time, what will be the distance between the two after 6 hours?

Options

A. a. `3sqrt(13)` km
B. b. `sqrt(39)` km
C. c. `sqrt(10)` km
D. d. `sqrt(127)` km

Answer

To solve the problem step by step:

### Step 1: Determine the distance traveled by Mohan
Mohan walks at an average speed of 1 km/hr. If he walks for 6 hours, the distance he covers can be calculated as:
\[ \text{Distance}_{\text{Mohan}} = \text{Speed} \times \text{Time} \]
\[ \text{Distance}_{\text{Mohan}} = 1 \, \text{km/hr} \times 6 \, \text{hrs} = 6 \, \text{km} \]

### Step 2: Determine the distance traveled by Sohan
Sohan walks at an average speed of 1.5 km/hr. If he also walks for 6 hours, the distance he covers can be calculated as:
\[ \text{Distance}_{\text{Sohan}} = \text{Speed} \times \text{Time} \]
\[ \text{Distance}_{\text{Sohan}} = 1.5 \, \text{km/hr} \times 6 \, \text{hrs} = 9 \, \text{km} \]

### Step 3: Visualize the positions of Mohan and Sohan
Mohan walks south while Sohan walks east. This creates a right triangle where:
- One leg (Mohan's distance) = 6 km (south)
- The other leg (Sohan's distance) = 9 km (east)

### Step 4: Apply the Pythagorean theorem
To find the distance between Mohan and Sohan, we can use the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Where:
- \( AB = 6 \, \text{km} \) (Mohan's distance)
- \( BC = 9 \, \text{km} \) (Sohan's distance)

Substituting the values:
\[ AC^2 = 6^2 + 9^2 \]
\[ AC^2 = 36 + 81 \]
\[ AC^2 = 117 \]

### Step 5: Calculate the distance \( AC \)
To find \( AC \), we take the square root of 117:
\[ AC = \sqrt{117} \]

### Step 6: Simplify \( \sqrt{117} \)
The square root of 117 can be simplified:
\[ \sqrt{117} = \sqrt{9 \times 13} = 3\sqrt{13} \]

### Final Answer
The distance between Mohan and Sohan after 6 hours is:
\[ \text{Distance} = 3\sqrt{13} \, \text{km} \]

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