Doppler effect on a police car?

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SUMMARY

The discussion centers on the Doppler effect as it applies to a police car's siren emitting a frequency of 60 Hz while moving at 50 m/s. The speed of sound is given as 300 m/s. The correct calculation for the frequency observed by a stationary observer behind the car is 51.4 Hz, derived from the formula f = 60 (300+0)/(300+50). The book's assertion of a frequency of 70 Hz, calculated using f = 60 (300+50)/(300+0), is incorrect in this context, as it misrepresents the scenario where the source is moving away from the observer.

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A police car's siren emits a wave with a frequency of 60 Hz. The speed of sound is 300 m/s and the car is moving 50 m/s. "what is the wavelength of the wave behind the car"

So, the answer seems pretty easy. The observer is a stationary point behind the car, and the source is moving away from the observer, which means frequency should decrease:

f = 60 (300+0)/(300+50) = 51.4 Hz
wavelength = 300/51.4 = 5.8

Is this logic correct?

However, the answer in the book states that the final frequency is 70 Hz, by this equation:
f = 60 (300+50)/(300+0) = 70 Hz

Which one is correct?
 
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An increase in the frequency would indeed make no sense at all. :)
 

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