Instanteous velocity

  1. 1. The problem statement, all variables and given/known data

    If an arrow is shot upward on the moon with a velocity of 59 m / s, its height in meters after t seconds is given by h = 59(t) - .83(t)². Find the instantaneous velocity after one second.


    2. Relevant equations




    3. The attempt at a solution

    h = 59(1) - .83(1)2
    h = 58.17

    h = 59(1.0000001)- .83(1.0000001)2
    h = 59.0000059 - 0.8300001660000083
    h = 58.1700057339999917

    v = 58.1700057339999917 - 58.17 / .0000001
    v = 57.339999917

    thats as far as i get. i think the instant vel would be 57.3 m / s because it wont take the answer 57 or 57.5.

    can someone help me?
     
  2. jcsd
  3. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF eplymale3043,

    Since you've posted this in the calculus forums, I'm guessing that we should use calculus. So, what is the definition of instantaneous velocity, or velocity in general?
     
  4. The speed at which an object is moving at any given time. right?
     
  5. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    Speed and velocity are two different quantities, but that's not what I was getting at. What is the definition of velocity in terms of distance and time?
     
  6. v =[tex]\Delta[/tex]d/[tex]\Delta[/tex]t
     
  7. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    Correct, that would be the average velocity. For instantaneous velocity one would take the limit (substituting d for h for clarity):

    [tex]v=\lim_{t\to0}\frac{\Delta h}{\Delta t} = \frac{dh}{dt}[/tex]

    Do you follow?
     
  8. Yes, so far.
     
  9. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    Now it's your turn to do some work. You are given h(t) in the question, all you need to do is find h'(t) and then plug in the numbers.
     
  10. h'(t) meaning the inverse?
     
  11. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member


    h'(t)
    meaning the first derivative of h with respect to t. As I said in my previous post, velocity is defined as the rate of change of displacement with time.
     
  12. I'm sorry, I dont understand what you mean.
     
  13. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    You are given h(t), which is a function describing how the height [displacement] of the arrow varies with time. Velocity is defined as the time derivative of displacement (height in this case), hence you need to take the derivative of h(t) with respect to t. In other words evaluate:

    [tex]\frac{dh}{dt} = \frac{d}{dt}\left(59t - 0.83t^2\right)[/tex]

    Does that make sense?
     
  14. Yes, that makes alot more sense.

    so, basically. I need to plug in the change in height for dh and change in time for dt.

    then solve? im still not sure what d by itself it.
     
  15. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    No, you need to take the derivative of h(t) and then plug in the numbers. You are studying calculus aren't you?
     
  16. Yes, but I'm really lost.

    How would I find the derivative?
     
  17. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    There are plenty of resources available on the internet regarding calculus, but I would suggest that you sit down a read the first few chapters on differentiation from your textbook. I'm more than happy to help you though this problem, but there is little more than can be done without knowledge of differentiation.
     
  18. After studying the chapter and doing some research in my calculus book, I have calculated that the velocity at 1 second is 57.34 m / s.

    Does this sound right to you?
     
  19. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    Sounds spot on to me :approve:
     
  20. yeah, it wanted me to keep finding the average velocity using. 1.5, 1.1, 1.05, 1.01, 1.005, and so on.

    basically, finding the limit.
     
  21. HallsofIvy

    HallsofIvy 40,229
    Staff Emeritus
    Science Advisor

    You need to back up and think about the course you are taking. If they are asking you to find the instantaneous velocity, they are assuming you know how to differentiate!
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook