Question:
A car is approaching a cliff at a constant speed. It sounds a horn when it is at 0.9 km from the cliff. The reflected sound of the horn is heard by the car driver after 5 sec. The speed of the car is: (Velocity of sound in air is \( 330 \text{ ms}^{-1} \))
A car is approaching a cliff at a constant speed. It sounds a horn when it is at 0.9 km from the cliff. The reflected sound of the horn is heard by the car driver after 5 sec. The speed of the car is: (Velocity of sound in air is \( 330 \text{ ms}^{-1} \))
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When a source moves toward a reflector, the distance the echo travels back to the source is effectively shortened by the displacement of the source during the sound's travel time.
Updated On: Jun 9, 2026
- \( 20 \text{ ms}^{-1} \)
- \( 30 \text{ ms}^{-1} \)
- \( 40 \text{ ms}^{-1} \)
- \( 50 \text{ ms}^{-1} \)
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The Correct Option is B
Solution and Explanation
Concept:
This problem involves the physics of sound wave reflection where both the source of the sound (the car) and the sound waves are moving. The time taken for the sound to travel to the cliff and back as an echo depends on the relative distance covered by the sound and the car during that interval.
Step 1: Define the variables and initial parameters.
Initial distance to the cliff \( d = 0.9 \text{ km} = 900 \text{ m} \). Time interval \( t = 5 \text{ s} \). Velocity of sound \( v_s = 330 \text{ ms}^{-1} \). Let the constant speed of the car be \( v_c \).
Step 2: Analyze the total distance covered.
In 5 seconds, the sound travels from the car to the cliff and reflects back to the driver. Distance covered by sound in 5 seconds = \( v_s \times t = 330 \times 5 = 1650 \text{ m} \). This distance is composed of the initial distance to the cliff (\( d = 900 \text{ m} \)) and the distance the sound covers on its way back to the driver, who has moved closer to the cliff.
Step 3: Set up the distance equation.
The sound travels \( 900 \text{ m} \) to reach the cliff. In the remaining time, the sound travels from the cliff to the car. The distance the car covers in 5 seconds is \( d_{car} = v_c \times 5 \). At the time the driver hears the echo, the distance of the car from the cliff is \( (900 - 5v_c) \). The total distance covered by sound is: $$ \text{Distance} = d_{to_cliff} + d_{from_cliff_to_driver} $$ $$ 1650 = 900 + (900 - 5v_c) $$ $$ 1650 = 1800 - 5v_c $$
Step 4: Solve for the velocity of the car \( v_c \).
$$ 5v_c = 1800 - 1650 $$ $$ 5v_c = 150 $$ $$ v_c = 30 \text{ ms}^{-1} $$ $$\boxed{30 \text{ ms}^{-1}}$$
Step 1: Define the variables and initial parameters.
Initial distance to the cliff \( d = 0.9 \text{ km} = 900 \text{ m} \). Time interval \( t = 5 \text{ s} \). Velocity of sound \( v_s = 330 \text{ ms}^{-1} \). Let the constant speed of the car be \( v_c \).
Step 2: Analyze the total distance covered.
In 5 seconds, the sound travels from the car to the cliff and reflects back to the driver. Distance covered by sound in 5 seconds = \( v_s \times t = 330 \times 5 = 1650 \text{ m} \). This distance is composed of the initial distance to the cliff (\( d = 900 \text{ m} \)) and the distance the sound covers on its way back to the driver, who has moved closer to the cliff.
Step 3: Set up the distance equation.
The sound travels \( 900 \text{ m} \) to reach the cliff. In the remaining time, the sound travels from the cliff to the car. The distance the car covers in 5 seconds is \( d_{car} = v_c \times 5 \). At the time the driver hears the echo, the distance of the car from the cliff is \( (900 - 5v_c) \). The total distance covered by sound is: $$ \text{Distance} = d_{to_cliff} + d_{from_cliff_to_driver} $$ $$ 1650 = 900 + (900 - 5v_c) $$ $$ 1650 = 1800 - 5v_c $$
Step 4: Solve for the velocity of the car \( v_c \).
$$ 5v_c = 1800 - 1650 $$ $$ 5v_c = 150 $$ $$ v_c = 30 \text{ ms}^{-1} $$ $$\boxed{30 \text{ ms}^{-1}}$$
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