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6 December, 08:24

The sides of a square baseball diamond are 90 feet long. When a player who is between second and third base is 60 feet from second base and heading towards third base at a speed of 22 feet per second, how fast is the distance between the player and home plate changing

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  1. 6 December, 08:51
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    d (L) / dt = 6,96 ft/s

    Explanation:

    We have a right triangle where the hypotenuse (L) is the distance between the player and home plate, the legs are the line between third base and home plate (90 feet) and distance between the player and third base (x) (over the line between second and third base). So we can write

    L² = (90) ² + x²

    Applying differentiation in relation to time, on both sides of the equation we have:

    2*LdL/dt = 0 + 2*x d (x) / dt (2)

    In this equation we know:

    d (x) / dt = 22 feet/sec

    x = 30 ft

    We need to calculate L when the player is at 30 feet from third base

    Then

    L² = (90) ² + (30) ²

    L² = 8100 + 900

    L = √9000

    L = 94,87 feet

    Then we are in condition for calculate d (L) / dt from the equation

    2*Ld (L) / dt = 0 + 2*x d (x) / dt

    2*94,87 * d (L) / dt = 2 * 30 * 22 ⇒ 189,74 d (L) / dt = 1320

    d (L) / dt = 1320 / 189,74

    d (L) / dt = 6,96 ft/s
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