A bat, moving at 5.25 m/s, is chasing a flying insect. The bat emits a 39.90-kHz chirp and receives back an ec

A bat, moving at 5.25 m/s, is chasing a flying insect. The bat emits a 39.90-kHz chirp and receives back an echo at 40.50 kHz. (Take the speed of sound in air to be v = 343 m/s.)

(a) What is the speed of the insect?

vr= v(fo-f)/(fo+f) ,(v=343m/s,fo=40.5kHz,f=39.9kHz)
vr= 343(40.5-39.9)/(40.5+39.9)= 2.5597m/s
vi= 5.25-2.5597= 2.6903= 2.69[m/s] (insect away speed)

check:
fd= 39.9(343-2.69)/(343-5.25)= 40.2KHz (insect hears)
fd= 40.2(343+5.25)/(343+2.69)= 40.5kHz (bat hears)