Question

Devanand’s house is 50 km West of Pradeep’s house. On Sunday morning, at 10 a.m., they leave their respective houses. Under which of the following scenarios, the minimum distance between the two would be 40 km?
Scenario I: Devanand walks East at a constant speed of 3 km/h and Pradeep walks South at 4 km/h.
Scenario II: Devanand walks South at 3 km/h and Pradeep walks East at 4 km/h.
Scenario III: Devanand walks West at 4 km/h and Pradeep walks East at 3 km/h.

• A. Scenario I only
• B. Scenario II only
• C. Scenario III only
• D. Scenario I and II
• E. None of the above

Hint:

For minimum-distance questions, set one point as origin, write the relative-position vector, minimize the quadratic \(d^2(t)\) (not \(d(t)\)) over \(t\ge0\).

Answer 1

The Correct Option is A

Place Pradeep at \((0,0)\) and Devanand at \((-50,0)\).

Scenario I

Velocities: Devanand \((3,0)\), Pradeep \((0,-4)\). Relative position of Devanand w.r.t. Pradeep after \(t\) hours: \[ \vec r(t) = (-50,0) + (3,0)t - (0,-4)t = (-50+3t,\,4t). \] Distance squared: \[ d^2(t) = (-50+3t)^2 + (4t)^2 = 2500 - 300t + 25t^2. \] This quadratic attains its minimum at \[ t = \frac{300}{2 \cdot 25} = 6 \text{ h}. \] Then \[ d^2(6) = 2500 - 1800 + 900 = 1600 \quad \Rightarrow \quad d_{\min} = 40 \ \text{km}. \]

Scenario II

Velocities: Devanand \((0,-3)\), Pradeep \((4,0)\). \[ \vec r(t) = (-50-4t,\,-3t), \quad d^2(t) = 25t^2 + 400t + 2500. \] The vertex occurs at \[ t = -\frac{400}{2 \cdot 25} = -8 < 0. \] Hence for \(t \ge 0\), \[ d_{\min} = d(0) = 50 \ \text{km}. \]

Scenario III

Velocities along the same line in opposite directions: \[ \vec r(t) = (-50 - 7t,\,0), \quad d(t) = |50 + 7t| \ge 50 \ \text{km}. \]

Conclusion

Only Scenario I yields minimum distance \(=40\) km.

\[ \boxed{\text{Scenario I only}} \]