How far is the bat from the wall when it hears the echo?

[ohms law]

Homework Statement

From Tipler & Mosca, Ch. 2, #56 (pg. 57)
Bats use echo location to place themselves. (I condensed all of this from the problem) So:
Bat flying at 19.5 m/s toward a vertical cliff.
Bat clicks.
Bat hears the echo 0.15 s later. (click travels at 343 m/s)

How close is the bat when it hears the echo?

Homework Equations

The position function for the bat is straightforward: $X_{bat}=19.5t$
the position function for the sound takes a little reasoning. The sound has to travel to the wall and back, so I'm going with $X_{click}={1/2} * 343t$.

The Attempt at a Solution

Solve these functions for t=0.15:
$X_{f bat}=19.5 m/s(0.15 s)=2.925 m$
$X_{click}={1/2} * 343 m/s(0.15 s)=25.725 m=X_{i bat}$

So, the bat's initial position is 25.725 m from the wall, and it's final position would be:
$X_{f bat}= X_{i bat} - X_{f bat} = 25.725 m - 2.925 m = 22.8 m$
So, the bat is 22.8 m from the wall when it hears the click.

 

[danielu13]

The way I did this was putting both velocities into one equation:

$\frac{1}{2}$*(v_Sound - v_Bat)*t

Using this, I got the answer of 24.263, which is the answer that the book gives for this question. The only thing that is different from your methodology is the fact that the $\frac{1}{2}$ affects v_Bat as well.

 

[ohms law]

Hey, thanks!
I didn't realize that the book gave an answer. I don't see one (and there isn't one in the SSM either).

Anyway... why would the velocity of the bat need to be halved? The sound has to travel to the wall and back, but the bat is traveling in a straight line (and at constant speed). So... wouldn't halving it's displacement mean that it hears the click instantaniously, when the click hits the wall?

 

[danielu13]

It doesn't, but I was lucky enough to find all of the solution manuals http://spotcos.com/misc/UW_PHYS_12X_ANSWERS/ while looking for help on a problem one night.

The reason for the velocity of the bat being halved is due to solving everything as one set that takes into account how long the entire distance is; the position of the bat factors in twice when dealing with the entire distance as a variable. If you download the solution manual for chapter 2 from that link, a step-by-step process is given for this method.

 

[ohms law]

Awesome, thanks for the link.