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A baseball diamond is a square 90 ft on a side. A player runs from first base to second base at 10 ft/sec. At what rate is the player's distance from home base increasing when he is half way from first to second base?

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5 answers
  1. Let the time after he leaves 1st base be t seconds
    Then the distance he has gone at that time be 10t ft from first, and we have a right-angled triangle with legs of 10t ft and 90 ft
    if the distance from the runner to home is d ft,
    d^2 = 90^2 + (10t)^2 = 8100 + 100t^2
    2d dd/dt = 200t
    dd/dt = 100t/d

    when he is halfway, 10t = 45
    t = 4.5 , and d^2 = 8100 + 100(4.5)^2 = 10125
    d = √10125 = 100.623 ft

    dd/dt = 100(4.5)/100.623 = appr 4.47 ft/s

    check my arithmetic

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    Reiny
  2. it didnt work out to be correct.

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  3. I checked the math

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  4. it had to have the square root in it still

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  5. geeeshh, then put the square root back in

    dd/dt = 100(4.5)/√10125 ft/sec

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    Reiny

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