posted by Sarah .
Suppose that a police car on the highway is moving to the right at 26 m/s, while a speeder is coming up from almost directly behind at a speed of 37 m/s, both speeds being with respect to the ground. The police officer aims a radar gun at the speeder. Assume that the electromagnetic wave emitted by the radar gun has a frequency of 6.00 x 10^9 Hz. Find the the difference between the frequency of the wave that returns to the police car after reflecting from the speeder's car, and the original frequency emitted by the police car.
F = ((V+Vr) / (V+Vs))*Fs.
F = ((343+37) / (343+26))*6*10^9 = 6.18*10^9 Hz.
F - Fs = 6.18*10^9 - 6.00*10^9 = 1.79*10^8 Hz.