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Two planes which are 2475 miles apart fly toward each other. Their

speeds differ by 75 mph. If they pass each other in 3 hours, what is
the speed of the slower plane?

User RegEdit
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1 Answer

3 votes

Final answer:

To find the speed of the slower plane, we set up an equation where the sum of the slower plane's speed and the faster plane's speed (which is greater by 75 mph) multiplied by 3 hours equals the distance between them, 2475 miles. Solving for the slower plane's speed gives us 375 mph.

Step-by-step explanation:

The question involves a scenario where two planes, initially 2475 miles apart, fly toward each other. Their speeds differ by 75 mph and they pass each other in 3 hours. We need to find the speed of the slower plane.

Let's denote the speed of the slower plane as s mph. Therefore, the speed of the faster plane would be s + 75 mph. Together, in 3 hours, they must cover the combined distance of 2475 miles which can be represented as:

3s + 3(s + 75) = 2475 miles

Solving the equation:

3s + 3s + 225 = 2475

6s = 2475 - 225

6s = 2250

s = 2250 / 6

s = 375 mph

Therefore, the speed of the slower plane is 375 mph.

User Kishan Solanki
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