Question

A robot is 4 m long and placed at a corner of a \(16\text{ m}\times 30\text{ m}\) rectangular field. It faces the diagonally opposite corner and reaches that corner in \(15\) s. What is its speed?

• A. 1 m/s
• B. 2 m/s
• C. 3 m/s
• D. 4 m/s

Hint:

When an object of length \(L\) starts with its rear at a point and moves toward a target point, the front travels (path length \(-\,L\)).

Answer 1

The Correct Option is B

Diagonal of the field \(= \sqrt{16^2+30^2}=\sqrt{1156}=34\ \text{m}.\)
Since the robot (length \(4\) m) is \emph{placed at the corner} with its back at the corner and front pointing along the diagonal, its nose already lies \(4\) m along the diagonal.
Distance the nose travels to reach the opposite corner \(=34-4=30\ \text{m}.\)
Time \(=15\) s, hence speed \(= \dfrac{30}{15}=2\ \text{m/s}.\) \[ \boxed{2\ \text{m/s}} \]