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Two jets leave an air base at the same time and travel in opposite directions. One jet travels 100 miles an hour faster than the other. If the two jets are 3924 miles apart after3 hours, what is the rate of each jet?

1 Answer

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Answer: speed of jet A is 704 miles per hour

speed of jet B is 604 miles per hour

Explanation:

Let the jets be jet A and jet B

Jet A and Jet B leave an air base at the same time and travel in opposite directions.

Let x = the speed of Jet A

Let y = the speed of jet B

One jet travels 100 miles an hour faster than the other. Let Jet A be the faster Jet. This means that

x = y + 100 - - - - - -1

If the two jets are 3924 miles apart after 3 hours, this means that both Jet A and Jet B travelled a total distance of 3924 miles after 3 hours.

Distance travelled = speed × time

Therefore,

Distance travelled by Jet A in 3 hours will be x × 3 = 3x miles.

Distance travelled by Jet B in 3 hours will be y × 3 = 3y miles.

Therefore, total distance is

3x + 3y = 3924 - - - - - - - -2

Substituting equation 1 into equation 2, it becomes

3(y+100) + 3y = 3924

3y + 300 + 3y = 3924

6y = 3924 - 300 = 3624

y = 3624/6 = 604 miles per hour

x = y + 100 = 604 + 100

x = 704 miles per hour

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