Answer
$36\frac{m}{s}$
Work Step by Step
We can calculate the observed frequency as
$f_{2}^{\prime}=(\frac{1}{1-\frac{u_2}{v}})f=\frac{124}{(1-\frac{26}{343})}=134.2Hz$
The speed of fast train can be determined as
$u_1=v(1-\frac{f}{f_{beat}+f_2^{\prime}})f$
We plug in the known values to obtain:
$u_1=(343)(1-\frac{124}{4.4+134.2})=36\frac{m}{s}$
