Understanding The Solution To A Doppler Effect Problem Can someone help me understand the solution to this problem? 

What I don't understand is 2 things:


*

*How has the radiation been Doppler shifted twice? I don't get what about the scenario leads to this being true.

*Why does being doppler shifted twice mean one has to multiply velocity/speed of light by 2 in order to find the change in frequency? 
 A: I will only give hints to solution. Actual calculation is up to you.
Consider two instants in emission, spaced $T$ in time ($T$ is wave's period). Call them $t_1$ and $t_2=t_1+T$. Follow both points of the wave, compute the times when they encounter the car, then follow them back after reflection. Let $t_1'$ and $t_2'$ the times when they reach speed camera again. $t_2'-t_1'$ is the received wave's period.
To give a further help: you may forget you are dealing with a wave. The problem is a purely kinematical one, concerning two particles always moving at speed $c$, forward and then backward.
A: The camera is the source of em radiation.
The detector adjacent to the camera “sees” an image of (the camera and) the transmitter as a result of the reflection of the em waves from the car.
That image is as far behind the car as (the camera and) the transmitter is in front of the car.
If the car is moving towards (the camera and) the detector at a speed $v$ what is the speed of the image of (the camera and) the transmitter relative to the detector?
There is your factor of two.
