...Can I record the sound of a passing car and determine the frequencies of approaching and leaving car ... and then just calculate the speed of the car on paper?

In principle yes, but the pre-requisite is that you need to know that the car itself is emitting the same sound on its approach and its departure. If it's an emergency vehicle with a siren, then that's a constant sound (or a constant range of tonal sweep), or if it's the car stereo turned up loud and you have perfect-pitch then that would also work. If you were relying on the engine noise then you'd require that the car is travelling at a constant speed (so the engine makes a constant-pitch hum/whine).

The frequency shift on the approach is f(heard) = f(made)*(1+v/c)

And on the departure is f(heard) = f(made)*(1-v/c)

where v is the velocity of the car and c is the speed of sound (330 metres per second = 740 mph).

There are 12 semitones in an octave (doubling of frequency), so each semitone (on an equally-tempered scale) is the 12th root of 2 fractional pitch change, ie frequency increases by 1.05946... (near enough 6%) going from one semitone to the next.

Since the before and after pitch-

*change* is 1+

**2***v/c, a pitch

*change* of one semitone would be equivalent to a speed of 3% of the speed of sound, ie 3% of 740mph ie 22mph.

A change of 2 semitones would be 44mph

3 semitones 66mph

4 semitones 88mph

5 semitones 110mph etc.