To solve this problem we will apply the concepts related to the Doppler effect. This is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically this is given as,
[tex]f = \frac{v \pm v_r}{v \pm v_s}(f_0)[/tex]
Here,
v = Speed of the waves in the middle
[tex]v_r[/tex] = Speed of the receiver in relation to the medium (Positive if the receiver is moving towards the transmitter or vice versa)
[tex]v_s[/tex] = Speed of the source with respect to the medium (Positive if the source moves away from the receiver or vice versa)
Our values are given as,
[tex]v = 342m/s[/tex]
[tex]f_0 = 262Hz[/tex]
[tex]v_r = 80m/s[/tex]
[tex]f = 350Hz[/tex]
Replacing,
[tex]350 = \frac{342+80}{342-v} (262)[/tex]
Solving for the velocity of the source,
[tex]v = 26.1m/s[/tex]
Therefore the speed of the other train is 26.1m/s
