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Instantaneous velocity animation

  1. Sep 2, 2004 #1
    A web page designer creates an animation in which a dot on a computer screen has a position of [tex]r^\rightarrow [4.40cm + (2.20cm/s^2)t^2]\underlinei + (5.00 cm/s)t\underlinej[/tex]

    okay i already have the correct answer, but i would like to know how the author got it. i came close to getting the correct answer.

    Question.) Find the instantaneous velocity at t=1.0. Give your answer as a pair of components separated by a comma(x,y).

    ok to find the instantaneous velocity, i need to find the derivative of the function...

    d/dt r = 2(2.20cm/s)*2(t)i + (5.00cm/s)
    plug in 1.0 for t and got....

    (8.8cm/s,5.00cm/s) <--- my answer

    (4.40cm/s, 5.00 cm/s) <--- correct answer

    it looks like the correct answer divided my x-component by 2, but why? can someone explain?
     
  2. jcsd
  3. Sep 2, 2004 #2

    Hurkyl

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    [tex]
    \vec{r} = [4.40cm + (2.20 cm/s^2)t^2] \hat{i} + (5.00 cm/s)t \hat{j}
    [/tex]

    That what you were trying to write?



    Anyways, you've identified that your answer is twice the alledgedly correct answer. I notice that your answer has several factors of two in it, so the first thought that springs to my mind is: "Can I find a reason why one of those 2's shouldn't be there?"
     
  4. Sep 2, 2004 #3
    I think...

    I think you are working the problem wrong. Take the derivative of the acceleration ONLY, we do not care about the initial position, we only want the derivative of a.

    [tex] \dfrac{d}{dt} 2.20t^2 [/tex]

    You do the math.

    Hope this helps.
     
  5. Sep 2, 2004 #4
    "2(2.20cm/s)*2(t)i"
    too many twos
     
  6. Sep 2, 2004 #5
    so...[tex](2.20cm/s^2)t^2[/tex] is the acceleration of i?

    wait...

    i got two 2's because you see 2.20cm/s^2? the derivative of that is 2(2.20cm/s) right? and the derivative of t^2 is equal to 2t. so...
    2(2.20cm/s) * 2t

    am i not suppose to care about the square root on cm/s?
     
  7. Sep 3, 2004 #6
    No...s^2 is simply a unit. It is NOT a variable. You are taking the derivative of the function with respect to the variable t, so try to envision that particular term as 2.20t^2...and the derivative of that would be 4.40t.
     
  8. Sep 3, 2004 #7

    Hurkyl

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    Doesn't matter if s is a unit or a varaible: it's a constant with respect to t, so its derivative (WRT t) is zero.
     
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