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32 votes
32 votes
an airplane is preparing to land at an airport. it is 48,000 feet above the ground and is descending at the rate of 3,200 feet per minute. at the same airport, another airplane is taking off and will ascend at the rate of 2,800 feet per minute. when will the two airplanes be at the same altitude and what will that altitude be?

User Rajnikant
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2 Answers

17 votes
17 votes

Final answer:

The two airplanes will be at the same altitude after 8 minutes of flight, at an altitude of 24,800 feet.

Step-by-step explanation:

To find the altitude at which the two airplanes will be at the same altitude, we need to find when their altitudes will be equal.

The descending airplane is decreasing its altitude at a rate of 3,200 feet per minute, so the equation for its altitude will be: A = 48,000 - 3,200t, where A is the altitude in feet and t is the time in minutes.

The ascending airplane is increasing its altitude at a rate of 2,800 feet per minute, so the equation for its altitude will be: A = 2,800t, where A is the altitude in feet and t is the time in minutes.

Setting these two equations equal to each other, we have: 48,000 - 3,200t = 2,800t

Solving for t, we get 48,000 = 6,000t. Dividing both sides by 6,000, we find t = 8 minutes.

Substituting this value of t back into either of the original equations, we can find the altitude at that time. Using the descending airplane equation, we have A = 48,000 - 3,200(8) = 24,800 feet.

User Zad
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2.3k points
22 votes
22 votes

Answer:

After 8 minutes ; altitude = 22,400 feet

Step-by-step explanation:

s1 = 2800t

s2 = -3200t + 48 000


2800t = -3200t + 48 000

28t = -32t + 480

7t = -8t + 120

7t + 8t = 120

15t = 120

15/15 t = 120/15

t = 8 minutes



s = 8 * (2800) = 22400 feet

User Masked Man
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3.1k points