Problema Solution
two recording devices are set 2600 feet apart, with the device at point A to the west of the device at point B. At the point on a line between the devices, 200 feet from point B, a small amount of explosives is detonated. The recording devices record the time the sound reaches each one. How far directly north of site B should a second explosion be done so that the measured time differences recorded by the devices is the same as that for the first detonation?
Answer provided by our tutors
let v = the speed of the sound
let E be the point where the first explosion was detonated then
AB = 2600 ft
AE = 2600 - 200 = 2400 ft
BE = 200 ft
since speed = distance/time => time = distance/speed
the measured time difference recorded by the device will be
2400/v - 200/v = (1/v)(2400 - 200) = (1/v)2200
let D be the point we want to determine directly north from B and let BD = x
the triangle ABD is a right triangle thus using the Pythagorean Theorem we can write for AD
AD^2 = AB^2 + BD^2
AD^2 = 2600^2 + x^2
the measured time difference in this case will be
AD/v - BD/v = (1/v)(AD - BD)
since the measured time differences need to be the same we have
(1/v)2200 = (1/v)(AD - BD) multiply both sides by v
AD - BD = 2200
AD = BD + 2200
(2600^2 + x^2)^0.5 = x + 2200
by solving we find
x = 436.4 ft
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the second explosion should be 434.4 feet directly north of site B.