Answer:
31.2 m/s
Step-by-step explanation:
= Frequency of approach = 480 Hz
= Frequency of going away = 400 Hz
= Speed of sound in air = 343 m/s
= Speed of truck
Frequency of approach is given as
eq-1
Frequency of moving awayy is given as
eq-2
Dividing eq-1 by eq-2
![(f_(app))/(f_(aw)) = (V + v)/(V - v)](https://img.qammunity.org/2020/formulas/physics/college/psrvh49rrlj0f2volh2plxw1u1cc865vrc.png)
![(480)/(400) = (343 + v)/(343 - v)](https://img.qammunity.org/2020/formulas/physics/college/q1vof1xglxbvita96g620gdz523fwuud1z.png)
= 31.2 m/s