A bat emits 50kHz pulse and receives 60kHz echo 200ms later?

Question

How mush distance now remains to the wall?
It says assume the speed of sound is 340 m/s.

Both frequencies are, of course, according to the bat's sensors.

The velocity of the bat is constant and he's heading directly into the wall.

Answer 1

bat's average distance = ct/2 = 34 m
Approximate solution for v (see ref. 1):
doppler delta-f/f = 2v/c
v = c(delta-f/f)/2 = 34 m/s
x0 (at time of transmission) = 34 m + vt/2 = 37.4 m
x1 (at time of reception) = 34 m - vt/2 = 30.6 m
This is only approximate because it applies to small v/c.
Exact solution for v (see ref. 2):
f1 (freq received/reflected by wall) / f0 = c/(c-v)
f2 (freq received by bat) / f1 = (c+v)/c
f2/f0 = (c+v)/(c-v)
v = c * (f2/f0 - 1) / (f2/f0 + 1) = 30.909 m/s
x0 (at time of transmission) = 34 m + vt/2 = 36.091 m
x1 (at time of reception) = 34 m - vt/2 = 31.909 m

Note that v(approx) approaches v(exact) as delta-f/f approaches zero.