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. \( 20 \text{ ms}^{-1} \)
• B. \( 30 \text{ ms}^{-1} \)
• C. \( 40 \text{ ms}^{-1} \)
• D. \( 50 \text{ ms}^{-1} \)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Picture the echo.
The car is $900\,\text{m}$ from the cliff when the horn sounds. The sound runs to the cliff, bounces back, and reaches the driver $5\,\text{s}$ later, by which time the car has crept closer.
Step 2: Total distance the sound travels.
Sound moves at $330\,\text{m s}^{-1}$ for $5\,\text{s}$, covering $330 \times 5 = 1650\,\text{m}$ in all (out to the cliff plus back to the moved car).
Step 3: Distance out to the cliff.
The forward trip to the cliff is the full $900\,\text{m}$.
Step 4: Distance back to the driver.
Let the car speed be $v_c$. In $5\,\text{s}$ the car advances $5v_c$, so when the echo returns the car sits $900 - 5v_c$ from the cliff. The return trip of the sound is therefore $900 - 5v_c$.
Step 5: Add the legs and equate.
\[ 900 + (900 - 5v_c) = 1650 \;\Rightarrow\; 1800 - 5v_c = 1650. \]
Step 6: Solve for the car's speed.
$5v_c = 1800 - 1650 = 150$, so $v_c = 30\,\text{m s}^{-1}$.
\[ \boxed{30\ \text{m s}^{-1}} \]