High School

Two jets leave an air base at the same time and travel in opposite directions. One jet travels 75 mi/h faster than the other. If the two jets are 13,023 miles apart after 9 hours, what is the rate of each jet?

Answer :

Final answer:

The rate of the slower jet is 686 mph, and the rate of the faster jet is 761 mph.

Explanation:

Let's denote the rate of the slower jet as x mph. Since the other jet travels 75 mph faster, we can denote the rate of the faster jet as x + 75 mph.

Since the two jets are traveling in opposite directions, their distances are adding up. After 9 hours, the total distance they have traveled is 13,023 miles. We can use the formula distance = rate * time to set up an equation:

  1. (rate of slower jet)(time) + (rate of faster jet)(time) = 13,023
  2. Substituting the values we have:
    x(9) + (x + 75)(9) = 13,023
  3. Simplifying the equation:
    9x + 9(x + 75) = 13,023
  4. Expanding and combining like terms:
    9x + 9x + 675 = 13,023
  5. Further simplifying:
    18x + 675 = 13,023
  6. Subtracting 675 from both sides:
    18x = 12,348
  7. Dividing both sides by 18:
    x = 686

Therefore, the rate of the slower jet is 686 mph, and the rate of the faster jet is 686 + 75 = 761 mph.

Learn more about opposite directions here:

https://brainly.com/question/11259650

#SPJ11

Other Questions