1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook