A car fitted with a device which transmits sound 60 times per minute. There is no wind and speed of sound in still air is 345 m/s. If you hear the sound 68 times per minute when you are moving towards the car with a speed of 12 m/s, the speed of the car must be nearly.

Options

A. 20.0 m/s towards
B. 30.0 m/s towards you
C. 10.0 m/s away from you
D. 10.0 m/s towards you

Explanation

To solve the problem, we will use the Doppler effect formula for sound. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given data
- The frequency of sound emitted by the car (f₀) = 60 times per minute = \( \frac{60}{60} \) Hz = 1 Hz
- The frequency of sound heard by the observer (f) = 68 times per minute = \( \frac{68}{60} \) Hz = \( \frac{17}{15} \) Hz
- Speed of sound in air (v) = 345 m/s
- Speed of the observer (v₀) = 12 m/s (moving towards the source)

### Step 2: Apply the Doppler Effect formula
The formula for the observed frequency when the source and observer are moving towards each other is given by:

\[ f = f₀ \cdot \frac{v + v₀}{v - v_s} \]

Where:
- \( f \) = observed frequency
- \( f₀ \) = emitted frequency
- \( v \) = speed of sound
- \( v₀ \) = speed of observer
- \( v_s \) = speed of source (car)

### Step 3: Substitute the known values into the formula
Substituting the known values into the Doppler effect formula:

\[ \frac{17}{15} = 1 \cdot \frac{345 + 12}{345 - v_s} \]

### Step 4: Simplify the equation
This simplifies to:

\[ \frac{17}{15} = \frac{357}{345 - v_s} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:

\[ 17(345 - v_s) = 15 \cdot 357 \]

### Step 6: Calculate the right side
Calculating the right side:

\[ 15 \cdot 357 = 5355 \]

So the equation becomes:

\[ 17(345 - v_s) = 5355 \]

### Step 7: Distribute the left side
Distributing gives:

\[ 5865 - 17v_s = 5355 \]

### Step 8: Isolate \( v_s \)
Rearranging to isolate \( v_s \):

\[ 5865 - 5355 = 17v_s \]

\[ 510 = 17v_s \]

### Step 9: Solve for \( v_s \)
Dividing both sides by 17:

\[ v_s = \frac{510}{17} = 30 \text{ m/s} \]

### Conclusion
Thus, the speed of the car must be nearly **30 m/s**.

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