The security alarm on a parked car goes off and produces a frequency of 960 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 95 Hz. At what speed are you driving?

Options

A. 9.2 m/s
B. 17 m/s
C. 12 m/s
D. 25 m/s

Answer

To solve the problem, we will use the Doppler effect equations for sound. The frequency observed by a moving observer when approaching and then moving away from a stationary source will be calculated.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency of the source, \( f_s = 960 \, \text{Hz} \)
- Speed of sound, \( v = 343 \, \text{m/s} \)
- Change in observed frequency, \( \Delta f = 95 \, \text{Hz} \)

2. **Doppler Effect Equations:**
- When the observer is moving toward the source:
\[ f_1 = \frac{v + v_c}{v} f_s \]
- When the observer is moving away from the source:
\[ f_2 = \frac{v - v_c}{v} f_s \]

3. **Calculate the Difference in Frequencies:**
- The difference between the frequencies when approaching and moving away is given by:
\[ f_1 - f_2 = 95 \, \text{Hz} \]
- Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{v + v_c}{v} f_s - \frac{v - v_c}{v} f_s = 95 \]

4. **Simplify the Equation:**
- Factor out \( \frac{f_s}{v} \):
\[ \frac{f_s}{v} \left( (v + v_c) - (v - v_c) \right) = 95 \]
- This simplifies to:
\[ \frac{f_s}{v} (2v_c) = 95 \]

5. **Rearranging for \( v_c \):**
- Rearranging gives:
\[ v_c = \frac{95v}{2f_s} \]

6. **Substituting Known Values:**
- Substitute \( v = 343 \, \text{m/s} \) and \( f_s = 960 \, \text{Hz} \):
\[ v_c = \frac{95 \times 343}{2 \times 960} \]

7. **Calculating \( v_c \):**
- Calculate the numerator:
\[ 95 \times 343 = 32585 \]
- Calculate the denominator:
\[ 2 \times 960 = 1920 \]
- Now divide:
\[ v_c = \frac{32585}{1920} \approx 16.95 \, \text{m/s} \]

8. **Final Result:**
- Rounding gives:
\[ v_c \approx 17 \, \text{m/s} \]

### Conclusion:
The speed at which you are driving is approximately **17 m/s**.