Answered

Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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

Sagot :

facundo3141592

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.

How to get the rate of each jet?

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:

https://brainly.com/question/4931057

#SPJ1

We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.