Question

A speeder is pulling directly away and increasing his distance from a police car that is moving at 24.0 m/s with respect to the ground. The radar gun in the police car emits an electromagnetic wave with a frequency of 6.95×10⁹ Hz. The wave reflects from the speeder's car and returns to the police car, where its frequency is measured to be 316 Hz less than the emitted frequency. Calculate the speeder's speed with respect to the ground.

Answer 1

here,

for electromagnetic waves there is no effect by the medium, like air.

So the only relevant velocity that counts is the velocity difference between sender and observer v,

i.e. the relative velocity of sender to observer.
The velocity is positive, when sender and observer move further apart as in this case. The general formula is
fobs = fsend(1 - v/c)

now when the observed wave is reflected, the reflected wave with reduced frequency is the sent wave, and its frequency will be reduced a second time.
So the frequency observed by the police car is

fobs = f refl (1 - v/c) = f send(1 -v/c)(1 - v/c)

fobs = f refl (1 - v/c) = f send(1 -v/c)(1 - v/c)

fobs = f send(1 - v/c)^2

6.95*10^9 - 316 = 6.95 * 10^9 * (1 - v/3*10^8)^2

v = 6.82 m/s

Since the police car went 24 m/s, the speed of the speeder is
24 + 6.82 = 30.82 m/s