Problema Solution
Two jets leave an air base at the same time and travel in opposite directions. One jet travels 83 mph slower than the other. If the two jets are 6005 mi apart after 5 hours, what is the rate of each jet?
Answer provided by our tutors
Let
v = the rate of the slower jet
v + 83 = the rate of the faster jet
d = 6005 mi the distance between them
t = 5 hr the time of the travel
Since speed=distance/time follows distance = speed*time
The planes are flying in opposite direction thus:
vt + (v + 83)t = d
v*5 + (v + 83)*5 = 6005
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v = 559 mph
v + 83 = 559 + 83 = 642 mph
The rate of the slower jet is 559 mph.
The rate of the faster jet is 642 mph.