Math

Plane A leaves the airport. One hour later, Plane B leaves the same airport on the same course. It catches up to Plane A in 2.5 hours. The average speed of Plane B is 300km/h faster than Plane A. Find the speed of each plane.

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  1. Recall that speed is distance traveled over time:
    v = d/t
    Represent the speeds with variables:
    Let x = speed of Plane A
    Since it was said the speed of Plane B is 300 km/h faster than Plane A,
    Let x + 300 = speed of Plane B

    When they catch up, that means that their distance traveled is the same, so we will equate their distance traveled.
    v = d/t
    d = v*t

    Plane A:
    d = x * (2.5 + 1)
    Note that the total time traveled by Plane A is 2.5 + 1 because it is one hour ahead of Plane B, and Plane B caught up with it after 2.5 hours.

    Plane B:
    d = (x + 300) * 2.5

    Equating,
    x * (2.5 + 1) = (x + 300) * 2.5
    3.5x = 2.5x + 750
    3.5 x - 2.5x = 750
    x = 750 km/h (Plane A speed)
    x+300 = 1050 km/h (Plane B speed)

    hope this helps~ `u`

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  2. Thanks so much! This helped a lot

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