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Math Help - calculus application

  1. #1
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    calculus application

    At 2.00pm the speedometer of your car read 30km/hr.At 2.10pm it reads 50km/hr.show that at sometime between 2.00pm and 2.10pm the acceleration is exactly 120km/hr^2.
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  2. #2
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    Quote Originally Posted by bobby77 View Post
    At 2.00pm the speedometer of your car read 30km/hr.At 2.10pm it reads 50km/hr.show that at sometime between 2.00pm and 2.10pm the acceleration is exactly 120km/hr^2.
    Let, v(t) be the velocity at and given time.
    So on the closed interval [2,2.1666] the intivial point is,
    v(2)=30
    The terminal point is,
    v(2.1666)=50
    Thus, the mean change is,
    \frac{50-30}{\frac{1}{6}}=120
    Hence, (assuming vecolity function is differenciable),
    By Lagrange's Theorem (Mean Value),
    There exists a point t\in (2,2.16666)
    Such as,
    v'(t)=a(t)=120
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