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Imagine that two planes leave Kansas City International Airport at different times. Plane A leaves the airport at 12 noon and heads north with a constant speed of 300 miles per hour. At 1:15, plane B leaves the airport and heads east with a speed of 400 miles per hour. At what time (on the clock) will the planes be 700 miles apart?

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Final answer:

Plane A travels 375 miles in 1 hour and 15 minutes. The remaining distance between the planes is 325 miles. Using the Pythagorean theorem, the distance covered by Plane B is calculated as 500 miles. The total time for the planes to be 700 miles apart is 2.5 hours, which corresponds to 2:30 PM.

Step-by-step explanation:

To find the time when the two planes will be 700 miles apart, we need to consider their respective distances and speeds. Plane A leaves the airport at 12 noon and travels at a constant speed of 300 miles per hour. Plane B leaves the airport at 1:15 PM and travels at a speed of 400 miles per hour.

By 1:15 PM, Plane A has already traveled for 1 hour and 15 minutes or 1.25 hours, which means it has covered a distance of 1.25 hours × 300 miles per hour = 375 miles.

To find the remaining distance between the two planes, we subtract the distance covered by Plane A from the total distance of 700 miles: 700 miles - 375 miles = 325 miles.

Since Plane B is moving perpendicular to Plane A, we can use the Pythagorean theorem to find the distance covered by Plane B: distance^2 = (325 miles)² + (375 miles)². Solving for distance gives us 500 miles.

Now, we can calculate the time taken by Plane B to cover the remaining distance: time = distance / speed = 500 miles / 400 miles per hour = 1.25 hours.

Adding the time taken by Plane A and Plane B gives us the total time: 1.25 hours + 1.25 hours = 2.5 hours.

Therefore, the planes will be 700 miles apart at 2.5 hours after Plane A's departure, which is 2:30 PM on the clock.

User Godeke
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