A sound detector at rest by a road sends a sound of 40.0 kHz towards a car that comes towards it at…

(a) The car as a listener. $f_1 = 40.0\ \text{kHz} \times \dfrac{340\ \text{m/s} + 25\ \text{m/s}}{340\ \text{m/s}} = 42.9\ \text{kHz}$.

(b) The car as a source. $f_2 = f_1 \times \dfrac{340\ \text{m/s}}{340\ \text{m/s} - 25\ \text{m/s}} = 40.0\ \text{kHz} \times \dfrac{365\ \text{m/s}}{315\ \text{m/s}} = 46.3\ \text{kHz}$.

(c) Beats. $46.35\ \text{kHz} - 40.0\ \text{kHz} = 6.35\ \text{kHz}$.

(d) The other car. The beats are $f\,\dfrac{v + u}{v - u} - f = f\,\dfrac{2u}{v - u}$. We want $u$, so from $5.00\ \text{kHz} \times (v - u) = 80.0\ \text{kHz} \times u$: