College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 12 - Problems - Page 465: 56

Answer

The distance between the bat and the moth is $0.337~m$

Work Step by Step

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 $0.337~m$
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