Two jets leave an air base at the same time and travel in opposite directions. One jet travels 80 mi/h faster than the other. If

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1 year ago

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the two jets are 11392 miles apart after 8 hours, what is the rate of each jet?

1 answer:

Vesna [10] · Answer 1 · 2024-01-25T18:06:48+03:00

Jet 1 moves with a speed of 672 mi/h, while jet 2 moves with a speed of 672 mi/h + 80mi = 752 mi/h.

Let's say that the rate (or speed) of the slower jet is R, then the rate of the faster jet is:

R + 80mi/h

Now, if we step on any of the two jets (such that we view it as if it doesn't move) the other jet will move with a speed equal to:

S = R + R + 80mi/h

We know that after 8 hours, the to jets are 11,392 mi apart, then we know that:

(R + R + 80mi/h)*8h = 11,392 mi

Now we can solve that for R:

2*R + 80mi/h = 11,392 mi/8h = 1,424 mi/h

R = 1,424 mi/h - 80mi/h)/2 = 672 mi/h

So Jet 1 moves with a speed of 672 mi/h, while jet 2 moves with a speed of 672 mi/h + 80mi = 752 mi/h.

If you want to learn more about speed:

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