Question

A body starts moving from the origin along a straight line at a speed of \(20\ \text{km h}^{-1}\) for \(1\) hour. Then it turns perpendicular to its path and moves with the same velocity for \(30\) minutes. Then the magnitude of its displacement is

• A. $10$ km
• B. $20$ km
• C. $10\sqrt{5}$ km
• D. $10\sqrt{2}$ km
• E. $20\sqrt{5}$ km

Hint:

For perpendicular motion: - Always use Pythagoras theorem - Treat distances as sides of a right triangle

Answer 1

The Correct Option is C

Concept: When motion occurs in perpendicular directions, displacement is found using Pythagoras theorem: \[ \text{Displacement} = \sqrt{x^2 + y^2} \]

Step 1: Calculate first displacement.
\[ \text{Speed} = 20\ \text{km h}^{-1}, \quad \text{Time} = 1\ \text{hour} \] \[ \text{Distance}_1 = 20 \times 1 = 20\ \text{km} \]

Step 2: Calculate second displacement.
\[ \text{Time} = 30\ \text{minutes} = \frac{1}{2}\ \text{hour} \] \[ \text{Distance}_2 = 20 \times \frac{1}{2} = 10\ \text{km} \]

Step 3: Apply Pythagoras theorem.
\[ \text{Displacement} = \sqrt{20^2 + 10^2} \] \[ = \sqrt{400 + 100} = \sqrt{500} = 10\sqrt{5}\ \text{km} \]