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Math Help - Instantaneous Velocity

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    Instantaneous Velocity

    If a ball is thrown straight up into the air with an initial velocity of 50_ft/s, its height in feet after t second is given by y = 50 t - 16 t^2. Find the average velocity for the time period beginning when t = 1 and lasting:

    (i) 0.1 seconds =16.4 i.e., over the time interval [1,1.1]
    (ii) 0.01 seconds=17.84

    (iii) 0.001 seconds=17.984

    Finally based on the above results, guess what the instantaneous velocity of the ball is when t=1 ?
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    Quote Originally Posted by qbkr21 View Post
    If a ball is thrown straight up into the air with an initial velocity of 50 ft/s, its height in feet after t second is given by y = 50 t - 16 t^2. Find the average velocity for the time period beginning when t = 1 and lasting:

    (i) 0.1 seconds =16.4 i.e., over the time interval [1,1.1]
    (ii) 0.01 seconds=17.84

    (iii) 0.001 seconds=17.984

    Finally based on the above results, guess what the instantaneous velocity of the ball is when t=1 ?
    It looks like we are to use the formula:
    \bar{v} = \frac{\Delta y}{\Delta t}

    So for i) we have that
    y(1) = 50 \cdot 1 - 16 \cdot 1^2 = 34 \, ft
    y(1.1) = 50 \cdot 1.1 - 16 \cdot 1.1^2 = 35.64 \, ft

    Thus
    \Delta y = y(1.1) - y(1) = 1.64 \, ft

    Thus
    \bar{v} = \frac{1.64 \, ft}{0.1 \, s} = 16.4 \, ft/s

    Similarly for ii) I get \bar{v} = \frac{y(1.01) - y(1)}{0.01 \, s} = 17.84 \, ft/s

    And for iii) I get \bar{v} = \frac{y(1.001) - y(1)}{0.001 \, s} = 17.984 \, ft/s

    It looks to me that the value of the instantaneous velocity at t = 1 s is 18 ft/s upward.

    -Dan
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    Bingo!! Dan thanks alot you are a Master
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