Answer:
31.2 m/s
Explanation:
[tex]f_{app}[/tex] = Frequency of approach = 480 Hz
[tex]f_{aw}[/tex] = Frequency of going away = 400 Hz
[tex]V[/tex] = Speed of sound in air = 343 m/s
[tex]v[/tex] = Speed of truck
Frequency of approach is given as
[tex]f_{app} = \frac{Vf}{V - v}[/tex] eq-1
Frequency of moving awayy is given as
[tex]f_{aw} = \frac{Vf}{V + v}[/tex] eq-2
Dividing eq-1 by eq-2
[tex]\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}[/tex]
[tex]\frac{480}{400} = \frac{343 + v}{343 - v}[/tex]
[tex]v[/tex] = 31.2 m/s
