A stationary police car emits a sound of frequency 1200 hz that bounces off a car on the highway and returns with a frequency of 1250 hz. the police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) how fast was the moving car going?

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Physics

Kiara Hart

17 February, 13:47

A stationary police car emits a sound of frequency 1200 hz that bounces off a car on the highway and returns with a frequency of 1250 hz. the police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) how fast was the moving car going?

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Villanueva · Answer 1

First, since the echo frequency is greater, the car must be roaming to the police car.

This is for the reason that the car is "running over" its own wave fronts, creating the wavelengths tinier, and henceforth increasing the frequency.

The trick here is to understand that the relative speed between the cars is half of the speed of the echo for the reason that the echo is returning to the police car at the similar speed as the sound wave trips to the moving car.

It would also appear that the speed of sound has been expected to be 337 m/s

So if v is the moving car's velocity, the correct formula to use is:

2v / 337 = (1250/1200) - 1

2v = 14.04 m/s

v = 7.02 m/s