Posted by Joel on November 04, 2002 at 18:12:16:
In Reply to: Re: Distance posted by MathBard on November 04, 2002 at 17:47:50:
: : A bullet is fired at a target. 2 seconds later the impact is heard. The speed of the bullet is 2500 feet per second and the speed of sound is 1100 feet per second. How far is the target?
: I bet one of the alpha-geeks can devise an intricate formula, but since the distance from A to B equals the distance from B to A, you can just average the speed (2500+1100)/2=1800 per second 1800ft to and 1800ft back takes 2 seconds, even though the bullet takes less than that to hit (.72 secs)and the sound more than that to return (1.28 secs).
... I thought you were the alpha-geek. :o)
Anyway, you can't do it that way.
You know that distance = speed x time.
So let t=time, d=distance, s=speed and if d = s * t, then
my "intricate" formula is just t = d/s
Here we have the bullet traveling to the target (distance d) at 2500 ft/sec so the time for that is:
t(1) = d/2500
Then, the sound travels back the same distance d at 1100 ft/sec, so the time for that is:
t(2) = d/1100
And we are given Total time T = 2 seconds, so
T = 2 = t(1)+t(2) = d/2500 + d/1100
2 = d(1/2500 + 1/1100)
d = 1527.7778 ft.
1527.7778/1100 = 1.3889 sec.
1527.7778/2500 = 0.6111 sec.
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