A person 'P' in train moving due south with 80 kmph and his friend Q moving due west at 60 kmph Q finds that P is travelling with a velocity of

Options

A. 100 kmph, `tan^(-1)(4/3)` S of E
B. 100 kmph, `tan^(-1)(4/3)` E of S
C. 100 kmph S of E
D. 100 kmph N of E

Answer

To solve the problem of finding the velocity of person P with respect to person Q, we can follow these steps:

### Step 1: Understand the Directions and Velocities
- Person P is moving due south with a velocity of 80 km/h.
- Person Q is moving due west with a velocity of 60 km/h.

### Step 2: Represent the Velocities as Vectors
- We can represent the velocity of P as a vector:
\[ \vec{v_P} = 0 \hat{i} - 80 \hat{j} \quad (\text{since south is negative } y \text{-direction}) \]
- The velocity of Q can be represented as:
\[ \vec{v_Q} = -60 \hat{i} + 0 \hat{j} \quad (\text{since west is negative } x \text{-direction}) \]

### Step 3: Calculate the Relative Velocity of P with Respect to Q
- The relative velocity of P with respect to Q is given by:
\[ \vec{v_{PQ}} = \vec{v_P} - \vec{v_Q} \]
- Substituting the vectors:
\[ \vec{v_{PQ}} = (0 \hat{i} - 80 \hat{j}) - (-60 \hat{i} + 0 \hat{j}) = 60 \hat{i} - 80 \hat{j} \]

### Step 4: Calculate the Magnitude of the Relative Velocity
- The magnitude of the relative velocity can be calculated using the Pythagorean theorem:
\[ |\vec{v_{PQ}}| = \sqrt{(60)^2 + (-80)^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \text{ km/h} \]

### Step 5: Determine the Direction of the Relative Velocity
- To find the direction, we can use the tangent function:
\[ \tan(\theta) = \frac{|\text{opposite}|}{|\text{adjacent}|} = \frac{80}{60} = \frac{4}{3} \]
- Therefore,
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]
- This angle is measured from the east towards the south, indicating that the direction of P's velocity relative to Q is south of east.

### Final Answer
- The velocity of person P with respect to person Q is:
- Magnitude: 100 km/h
- Direction: \( \tan^{-1}\left(\frac{4}{3}\right) \) south of east.