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

Question

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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Vodka 3 years 2021-08-06T17:30:57+00:00 1 Answers 31 views 0

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  1. linhdan
    0
    2021-08-06T17:32:45+00:00
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    Answer:

    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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