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Calculate the instantaneous acceleration

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    A car's velocity as a function of time is given by Vx (t) = alpha + beta (t)^2, where alpha = 3.00m/s and beta = .100m/s^3.

    Calculate the instantaneous acceleration for t= 5seconds
    2. Relevant equations

    ax = dv/dt


    3. The attempt at a solution

    My first attempt is to get various times and get their corresponding acceleration.

    vx @ 0s = 3m/s
    vx @ 1s = 3.1m/s
    vx @ .5s = 3.025 m/s
    vx@ 5s = 5.5

    I know compared a= 3.1-3/1-0 = .1
    3.035-3/.5-0 = .05

    the average would end up being around .1

    I know this problems involved more calculus math, derivitives etc.

    How do I apply this here.
     
  2. jcsd
  3. Sep 9, 2007 #2
    Lets go over a few easy rules for calculus:

    [tex]\frac{d}{dx}c = 0[/tex] Constants always become zero when take a derivative

    [tex]\frac{d}{dx}x^n = nx^{n-1}[/tex] The power rule, if there is a coefficient multiply n by the coefficient.

    But if you don't really know calculus you probably shouldn't worry about it...

    If you want to approximate the instantaneous acceleration consider this: the idea of an instantaneous rate of change is finding an average between two points secant line, and shrinking the distance between those points until they are infinitely small right? This forms what we call a tangent line. Since you don't have a way to do this without calc you could do the next best thing find really small averages from t=5s. I'm talking .1, .01, .001, .0001 off from 5. You don't want to do 0, 1, .5, and 5 -- None of these are really very close at all.
     
  4. Sep 9, 2007 #3
    yeah but that will give me .00001

    is this correct?

    also, when I chose close numbers to the time would it not give me an number near one alway?
     
    Last edited: Sep 9, 2007
  5. Sep 9, 2007 #4
    what is the derivative of your velocity equation?
     
  6. Sep 9, 2007 #5
    well the equation would look like this (keep in mind I am taking calc along with physics):

    3.00 + .100 t^2

    which leads to:

    3+.2 = 3.2?

    I don't know much about derivitives yet...
     
  7. Sep 9, 2007 #6
    i'd be real worried then ... :O taking the derivative is real ez, i would go ahead and scan your calculus book on how to do so.

    the derivative would be

    [tex]a_x(t)=2\beta{t}[/tex]
     
    Last edited: Sep 9, 2007
  8. Sep 9, 2007 #7
    I'm taking Physics with Calc, Calc being taken at the same time, now about the problem.o:)
     
  9. Sep 9, 2007 #8
    if you want some examples i'll be more than happy to show you.
     
  10. Sep 10, 2007 #9
    No, when I mean find the average over a small interval I mean

    [tex]m = \frac{f(x_2)-f(x_1)}{x_2-x_1} = \frac{v_{x}(5+h)-v_{x}(5)}{(5+h)-5}[/tex]

    where h is very very small...

    Or better yet if you know how, take the limit as h -> 0
     
  11. Sep 10, 2007 #10

    [tex]m = \frac{v_{x}(5+.00001)-v_{x}(5)}{(5+.00001)-5}[/tex]?

    = .00005/.00005 = 1?

    nvm, I think I got it, I am working on it, I forgot to plug into the original equation provided.
     
    Last edited: Sep 10, 2007
  12. Sep 10, 2007 #11
    Update:

    5.5 @ 5s
    5.50001 @ 5.00001s

    5.5 - 5 / 5.00001 - 5 = 50000m/s^2 ??? Is this right, seems large :redface:
     
    Last edited: Sep 10, 2007
  13. Sep 10, 2007 #12

    learningphysics

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    1 is the right answer.
     
  14. Sep 10, 2007 #13
    Is is really, I marked 1 again, but it tells me I am incorrect. :(

    I got two more tries, I don't know if I shuold mark 1.00 (but isn't that more precise)?

    lol
     
  15. Sep 10, 2007 #14

    learningphysics

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    Make sure you have posted the question right...
     
  16. Sep 10, 2007 #15
    lol, ok, (I was marking it on the wrong question) :blushing:

    So the way I did it earlier would be the correct way to solve for it. Let's say time is 0. Then it would be:

    (0+.00001) - 0 / (0+.00001 - 0) = 1

    If this is right, then all instant acceleration would have to be 1 . :uhh:
     
  17. Sep 10, 2007 #16

    learningphysics

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    Yes, you can do it that way... that will give you a good approximation... for exact results you need derivatives.
     
  18. Sep 10, 2007 #17
    The point I am trying to make is that regardless of time, will the instant acceleration alway be one. Because if you add a number to the original time (ex: .0000000001) and follow the equation format, then you will eventually end up with the same thing =1.

    But in the question I am trying to solve, when I place 1 as the answer when time = 0 it would mark it as incorrect.

    BTW, I just remembered that the top is going to be (3+.00001)- 0 / (0+.00001-0)= 1.

    :cry:
     
  19. Sep 10, 2007 #18

    learningphysics

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    V(0) = 3
    V(0.0001) = 3.000000001

    [V(0.0001) - V(0)]/(0.0001 - 0) = 0.000000001/0.0001 = 0.00001 not 1.
     
    Last edited: Sep 10, 2007
  20. Sep 10, 2007 #19

    learningphysics

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    No at t=0, it is not 1. It is only 1 at t = 5.
     
  21. Sep 10, 2007 #20

    learningphysics

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    From rocophysics' post

    [tex]a_x(t)=2\beta{t}[/tex]

    where [tex]\beta = 0.1[/tex]
     
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