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!

Calculus: Differentiation

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data

    The height "s" at time of a silver dollar dropped from a building is given by s(t) = -16t^2 + 1350, where "s" is measured in feet and "t" is measured in seconds
    [s'(t) = -32t]

    a) Find the average velocity on the interval [1,2].
    ( I ALREADY SOLVED)
    b) Find the instantaneous velocity when t=1 and t=2. (I ALREADY SOLVED)
    c) How long will it take to hit the ground?

    Not sure about this one. Here's what I did: t=sqrt of 1350/16 ≈ 9.2

    d) Find the acceleration of this object at any given point in time.


    Not sure about this one. Here's what I did:
    v(t) = s'(t) = -32t.

    e) What's the velocity of this object when it hits the ground?
    (Really have no idea how to find it)

    Thanks in advance! I really need to learn how to do this type of problem for my upcoming quiz.
     
  2. jcsd
  3. Sep 29, 2012 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, this is right. Can you explain how you got it?
    OK, that's the velocity as a function of time. How are acceleration and velocity related?
    In part (c), you found the time at which the object hits the ground. And you also have the velocity as a function of time. So how can you combine these facts?
     
  4. Sep 29, 2012 #3
    For d) Find the acceleration of this object at any given point in time.
    I simply followed what the book states:

    "If s is the position function for an object moving along a straight line, then the
    acceleration of the object at time t is given by a(t) = v' (t)"

    Here's my calculation:
    V'(t) = -16(t+Δt)^2 + 1350 - (-16t^2 + 1350)/ Δt

    the answer was -32t.

    Thank you for assisting me!
     
    Last edited: Sep 29, 2012
  5. Sep 29, 2012 #4
    e) what's the velocity of the object when it hits the ground?

    "In part (c), you found the time at which the object hits the ground. And you also have the velocity as a function of time. So how can you combine these facts?"

    Could it be time x velocity = (9.2)(-32) = 294.4ft/sec ?

    Thank you for helping :)
     
  6. Sep 29, 2012 #5

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    But earlier, you said that v(t) = s'(t) = -32t, which is correct.
    Now what is a(t) = v'(t)? You have to differentiate again.
     
  7. Sep 29, 2012 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, the answer is right but it's not "time x velocity". It is velocity AT time 9.2, which is v(9.2). And since v(t) = -32t, it follows that v(9.2) = -32*9.2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculus: Differentiation
  1. Differential Calculus (Replies: 10)

  2. Differential calculus (Replies: 2)

Loading...