Answer
The train is moving with a velocity of 25.8 m/s
Work Step by Step
We first write an expression for the frequency $f_a$ heard as the train approaches.
$f_a = \frac{f}{(1-\frac{v_{train}}{v_{snd}})}$
$f = f_a~(1-\frac{v_{train}}{v_{snd}})$
We then write an expression for the frequency $f_r$ heard as the train recedes.
$f_r = \frac{f}{(1+\frac{v_{train}}{v_{snd}})}$
$f = f_r~(1+\frac{v_{train}}{v_{snd}})$
Next, we equate the two expressions for $f$ and solve for $v_{train}$:
$f_a~(1-\frac{v_{train}}{v_{snd}}) = f_r~(1+\frac{v_{train}}{v_{snd}})$
$(\frac{v_{train}}{v_{snd}})(f_r+f_a) = f_a-f_r$
$v_{train} = \frac{(f_a-f_r)(v_{snd})}{f_r+f_a}$
$v_{train} = \frac{(565~Hz-486~Hz)(343~m/s)}{486~Hz+565~Hz}$
$v_{train} = 25.8~m/s$
The train is moving with a velocity of 25.8 m/s.
