Question

An aeroplane is flying with a uniform speed of \( 150 \text{ km h}^{-1} \) along the circumference of a circle. The change in its velocity in half the revolution (in km h\(^{-1}\)) is

• A. $150$
• B. $100$
• C. $200$
• D. $300$
• E. $50$

Hint:

Half revolution → velocity reverses → change = $2v$ (maximum possible).

Answer 1

The Correct Option is D

Concept: Velocity as a vector quantity in circular motion
Velocity is a vector, meaning it has:
• Magnitude (speed)
• Direction Even if speed remains constant, velocity changes whenever direction changes. ---

Step 1: Understand the motion
The aeroplane is moving in a circular path with: \[ \text{Speed} = 150 \text{ km h}^{-1} \] In circular motion:
• Direction of velocity is always tangential
• Magnitude remains constant ---

Step 2: Define initial velocity vector
Let initial velocity be: \[ \vec{v}_1 = 150 \text{ km h}^{-1} \] Direction: tangent at starting point ---

Step 3: After half revolution
After half revolution:
• Plane reaches opposite point on circle
• Direction of motion reverses Thus: \[ \vec{v}_2 = -150 \text{ km h}^{-1} \] ---

Step 4: Compute change in velocity
\[ \Delta \vec{v} = \vec{v}_2 - \vec{v}_1 \] \[ = (-150) - (150) = -300 \] ---

Step 5: Magnitude of change
\[ |\Delta \vec{v}| = 300 \text{ km h}^{-1} \] ---

Step 6: Vector interpretation (important)

• Initial and final velocities have equal magnitude
• Directions are exactly opposite
• Angle between them = $180^\circ$ Using vector formula: \[ |\Delta v| = \sqrt{v^2 + v^2 - 2v^2 \cos 180^\circ} \] \[ = \sqrt{v^2 + v^2 + 2v^2} = \sqrt{4v^2} = 2v \] \[ = 2 \times 150 = 300 \] ---

Step 7: Physical understanding

• Maximum possible change in velocity occurs when direction reverses
• This happens at half revolution in circular motion --- Final Answer: \[ \boxed{300} \]