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Homework Help: Applications of Derivative

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data
    The height of an object moving vertically is given by
    s = -16t2 + 96t + 112,
    when s in ft and t in sec.
    a. Find the object's velocity when t=0
    b. Find its maximum height and when it occurs.
    c. Find its velocity when s=0

    3. The attempt at a solution

    a.
    Now I know to find the objects velocity, take the derivative, which is:

    v(t) = s' = -32t + 96.
    when t=0, the objects velocity is 96 m/s

    b.
    This is when I got stuck. When the velocity is 96 m/s, what is its maximum height. I am guessing set the derivative's velocity equal to 0, and solve for t. and plug the value of t in the anti derivative equation, and you would get the maximum height, is the correct? Because whenever the derivative is zero, the anti derivative would either be a max or min, in this case, a max.

    0 = -32t + 96
    t = 3 sec

    s = -16t2 + 96t + 112 plug t= 3 in this equation and you will get,
    s = 256 ft?

    c.

    With the assumptions that my work is so far correct, this is what I would do. set the anti derivative equal to zero and solve for t, using factorization.

    0 = -16t2 + 96t + 112
    0 = -16(t2 -6t - 7)
    0 = -16(t + 1)(t - 7)
    t = -1, t = 7, because those are the values when the height is zero.

    I would plug those values into the derivative to find the velocity at that point.

    v(-1) = 128 m/s
    v(7) = -128 m/s

    Is all my work correct? Or did I made some mathematical or logical errors?
     
    Last edited: Mar 21, 2010
  2. jcsd
  3. Mar 21, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Your answers are correct, except you used the wrong units in a).
    Also in b, you say "When the velocity is 96 m/s, what is its maximum height.". That statement is nonsense, and you don't need it. The 96 m/s was about a specific time instance (t = 0), now we're talking about a different time instance (namely, t for which v is maximal) which you find by setting s' = 0.

    Finally, I want to make a remark - perhaps a bit picky - about your use of the word "anti derivative". Personally, I would call s "the (distance / given / original) function", and v its derivative. If v was given, I would call that the given/original function and refer to s as the anti-derivative. Now when you say "anti-derivative" it may mean that you are looking at a function f, such that the derivative f' = s (and f'' = s' = v).
     
  4. Mar 21, 2010 #3

    Thank you so much.

    And it was interesting how I never thought about the anti-derivative, like what it may mean when not used appropriately, thank you.
     
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