Answer
$19.02\space m/s$
Work Step by Step
Here we use the equation $f^{'}=(\frac{f}{1\space \pm\space u/V })$ Where $f^{'}$ - shifted frequency, f - original frequency, u - speed of the car, V - speed of the sound wave in the air.
$f^{'}=(\frac{f}{1\space \pm\space u/V })$ We'll use the minus sign because the source is approaching.
$f^{'}=(\frac{f}{1\space-\space u/V })$
$1-\frac{u}{V}=\frac{f}{f^{'}}$
$1-\frac{f}{f^{'}}=\frac{u}{V}$
$u=V(1-\frac{f}{f^{'}})$ ; Let's plug known values into this equation.
$u=343\space m/s(1-\frac{494Hz}{523Hz})=19.02\space m/s$
