physics
posted by Liz .
A speeding car is pulling away from a police car. The police car is moving at 30 m/s. The radar gun in the police car emits an electromagnetic wave with a frequency of 20.0 x 10^9 Hz. The wave reflects from the speeding car and returns to the police car where the frequency is measured to be 4750 Hz less than what was emitted. The speed of light is 3 x 10^8 m/s. How fast is the speeding car going?
I know you have to set up two equations (one for before the bounce off the speeding car and one after), but I can't seem to work it to a reasonable number.
The equation is:
fd=(v±vd)/(v±vs)fs
where fd= frequency received by a detector
fs= frequency of source
v= velocity of the wave
vd= velocity of the detecor
vs= velocity of the source
I've been working on it for days, but the best thing I've gotten was that the car was traveling at 101 m/s, which doesn't sound possible to me lol. Could anyone look at this and explain to me what I'm doing wrong?

The equation you have been given is actually not correct for light and radar guns, but it is close enough. For velocities much less than that of light, there is a frequency shift nthat is more easily written as just
deltaf/f = 2*V/c
The frequency shift is deltaf. The factor of two is the result of the echo doubling the Doppler shift. The signal received by the car is shifted delta f/f and the reflected signal is twice that because the car is moving away from the receiver (radar gun).
The formula you have been told to use should give the same result, to three significant figures or more, if used correctly.
The exact relativisitic Doppler formula for light can be found at
http://en.wikipedia.org/wiki/Relativistic_Doppler_effect
In your case
deltaf/f = 4750/2*10^10 =
= 2.38*10^7 = 2 V/(3*10^8)
V = 35.6 m/s = 128 km/h 
I actually used that formula and got that answer (35.625 m/s) but does that take into account the speed of the police car? (or is that a red herring)