Answer
(a) moving away
(b) $25.09m/s$
Work Step by Step
(a) According to the Doppler Effect $f^{\prime}=f(1\pm \frac{u}{c})$. The radar beam which was reflected from the speeding car was lower in frequency than the original beam. Therefore, from eq(1) the source was receding away and thus the speeding car was moving away from the highway.
(b) We know that
$f-f^{\prime \prime}=\frac{2u}{c}f$
We plug in the known values to obtain:
$4.04\times 10^3Hz=\frac{2u}{3\times 10^8m/s}(24.150\times 10^9Hz)$
This simplifies to:
$u=25.09m/s$
