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Math Help - Rate of change help

  1. #1
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    Rate of change help

    The height (in m) that an object has fallen from a height of 180 m is given by the position function s(t) = -5tsquared + 180, where t > or equal to 0 and t is in seconds.

    a. Find the average velocity during each of the first 2 seconds.

    b. Find the velocity (instantaneous) of the object at t = 4

    c. At what velocity will the object hit the ground?

    I tried these, and got different answers then the correct ones. Help me out?
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  2. #2
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    Please SHOW your work.

    Velocity at a moment? That's what a derivative is for.

    Average Volocity over time? That's what integrals are for.

    When does it hit the ground? That's what the position function is for.

    Let's see how you managed what you managed.
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  3. #3
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    Quote Originally Posted by Hockey_Guy14 View Post
    The height (in m) that an object has fallen from a height of 180 m is given by the position function s(t) = -5tsquared + 180, where t > or equal to 0 and t is in seconds.

    a. Find the average velocity during each of the first 2 seconds.

    b. Find the velocity (instantaneous) of the object at t = 4

    c. At what velocity will the object hit the ground?

    I tried these, and got different answers then the correct ones. Help me out?
    a) The average velocity between times t_1 and t_2 is found by the formula \frac{s_2 - s_1}{t_2 - t_1}.

    b) The instantaneous velocity can be found by taking the first time derivative of the position function s(t).

    c) When does the object hit the ground? When s(t) = 0, of course. So solve for t and plug that into the velocity equation you got for b.

    -Dan
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