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Calculus Rocket Problem (Derivatives)

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A model rocket is launched vertically upward. If it's position (height) y above the ground in feet as a function of time t in seconds is given by:

    2. Relevant equations

    y = f(t) = 192t - 16t^2

    3. The attempt at a solution

    1. Find the instantaneous velocity and acceleration at any time t.

    Now this is my first calculus course on my long academic journey, but velocity is the first derivative and acceleration is the second.

    Velocity -

    y' = 192-32t

    That was easy, but it is the second derivative (acceleration) that is stumping me:

    y'' = -32

    How is that possible.. negative acceleration the entire time?

    Thank you for your assistance.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 4, 2008 #2


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    Science Advisor

    That's not much of a rocket! If y= 192t- 16t2, then the "rocket" provides an (instantaneious!) initial 192 ft/sec speed and then shuts down. The only force acting on it is that due to gravity. Are you sure that the y= 192t- 16t2 isn't from thepoint at which the rocket shuts off?
  4. Mar 4, 2008 #3
    That is the equation, verbatim from the homework problem.

    One of the other questions is (and I have done the work for it):

    2. How high will the rocket reach?

    192 - 32t = 0

    -32t = -192
    -32 = -32

    total time = 6 seconds

    192(6) - 16(36) = 576 feet
  5. Mar 4, 2008 #4
    That's the correct approach. Plug your critical point back into your equation, and you find the Y_max!
  6. Mar 4, 2008 #5
    Thanks roco, but is the second derivative (acceleration) correct at -32? It just does not seem logical to me.
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