## College Physics (4th Edition)

The distance between the bat and the moth is $0.337~m$
We can find the speed of sound in the air: $v = 331+0.6~T = 331+(0.6)(10.0^{\circ}C) = 337~m/s$ Since the speed of sound is much greater than the speed of the bat or the moth, we can ignore the slight change in position of the bat and the moth during the 2.0-ms time interval. We can find the total distance that the sound wave travels in $2.0~ms$: $d = v~t = (337~m/s)(2.0\times 10^{-3}~m/s) = 0.674~m$ Since the distance between the bat and the moth is half the total distance the sound travels, the distance between the bat and the moth is $\frac{0.674~m}{2}$ which is $0.337~m$