Question

A vehicle starts from rest and accelerates along straight path at \( 2 \, \text{m/s}^2 \). At the starting point of the vehicle, there is a stationary electric siren. How far has the vehicle nearly gone when the driver hears the siren at 94% of its value when the vehicle was at rest? (speed of sound = 220 m/s)

• A. 98 m
• B. 49 m
• C. 196 m
• D. 24.5 m

Hint:

The Doppler effect is useful for understanding the change in frequency when the observer or source is moving. The observed frequency increases when the observer moves towards the source.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This is a Doppler Effect problem where the observer (driver) is moving away from a stationary source. We find the velocity required for the frequency shift and then the distance traveled under constant acceleration.
Step 2: Key Formula or Approach:
1. Doppler shift: \( f' = f \frac{v - v_o}{v} \).
2. Kinematics: \( v^2 = 2as \).
Step 3: Detailed Explanation:
Given: \( f' = 0.94 f \), \( v = 220\text{ m/s} \), \( a = 2\text{ m/s}^2 \).
\[ 0.94 f = f \left( \frac{220 - v_o}{220} \right) \]
\[ 0.94 = 1 - \frac{v_o}{220} \implies \frac{v_o}{220} = 0.06 \]
\[ v_o = 220 \times 0.06 = 13.2\text{ m/s} \]
Now, find the distance \( s \) using kinematic equation:
\[ v_o^2 = u^2 + 2as \implies (13.2)^2 = 0 + 2(2)s \]
\[ 174.24 = 4s \implies s = \frac{174.24}{4} = 43.56\text{ m} \]
The nearest value in the options is 49 m.
Step 4: Final Answer:
The vehicle has gone nearly 49 m.