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Car moving in a circular motion

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    A car is moving along a 4km long circular track with constant tangential acceleration. If it starts from rest and it does 2km between 2s and 8s find the vector acceleration, the velocity vector and the displacement after 16s


    2. Relevant equations



    3. The attempt at a solution
    All right, I am not sure about this, but I gave it a shot.

    So, reading the problem e understand that the car did 2000meters in 6seconds.
    So I'd say that:

    [itex]x_f = 1/2 at^2[/itex]
    [itex]2000 = 1/2a(6)^2 \rightarrow 18a = 2000 \rightarrow a = 111m/s^2[/itex]

    Since I am a bit clueless, I think what I just did is calculating the value of the acceleration, but obviously not the vector.
    Now, uhm. As long as I understood to get the acceleration vector I need to sum up radial and tangential acceleration, but I am just stuck, could someone help me out?
     
  2. jcsd
  3. Dec 3, 2011 #2

    Doc Al

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    Here you are working on finding the tangential acceleration. Unfortunately, your equations assume it went 2000 m starting from rest, but that 2000 m is traveled between 2 s and 8 s. It's already moving fast at the 2 second mark.

    Hint: You'll need two equations to solve for the tangential acceleration.

    To find the total acceleration, you'll also need the radial acceleration. How will you find that?
     
  4. Dec 3, 2011 #3
    So the vector acceleration is given by...

    A(vec) = aradial(vec) + atangential(vec)

    the tangential one is given by dv/dt.
    While the radial one is given by [itex]\frac{-v^2}{r}[/itex]
     
  5. Dec 3, 2011 #4

    Doc Al

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    Yes. Looks like you've got it. (That minus sign in the radial acceleration just means that it is directed towards the center. It's the actual direction that counts.)
     
  6. Dec 3, 2011 #5
    But I am not exactly sure what I am supposed to plug in in the formulas.
    Because, I don't know the initial speed, I just know that between 2 and 8 seconds it goes 2km with constant acceleration.
    Meaning I can't use
    xf = xi + v0t +1/2at^2 because I don't have the initial velocity nor the acceleration. >.<
     
  7. Dec 3, 2011 #6

    Doc Al

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    Don't give up so easily. :wink:

    That's one of the equations you'll need. What's another? You'll need two equations, since you have two unknowns.
     
  8. Dec 3, 2011 #7
    Of all the other kinematics I can think of
    [itex]v^2_f = v^2i + 2a(x_f - x_i)[/itex]
    But I don't have the initial velocity, the final velocity and the acceleration >.<

    I was thinking about some circular equation, maybe?
     
  9. Dec 3, 2011 #8

    Doc Al

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    What about relating speed and time?
     
  10. Dec 3, 2011 #9
    you mean :
    xf = xi+vt ?
    But can't we use this just when we have constant speed?
     
  11. Dec 3, 2011 #10

    Doc Al

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    Not that one. Choose one that relates velocity with time.
     
  12. Dec 3, 2011 #11
    Ohhh I got it! xf = xi+1/2(vi + vf)t
    Now I should be able to solve for the two unknowns :)
     
  13. Dec 3, 2011 #12
    But using this one, I still have three unknowns, don't I? The initial and final velocity plus the acceleration!
    @_@ I'm getting confused. Maybe I have to add the equation I wrote before and do a system of three equations in three unknowns?
     
  14. Dec 3, 2011 #13

    Doc Al

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    Choose another equation! Hint: What's the definition of acceleration?
     
  15. Dec 3, 2011 #14
    accelerationi is defined as the change of velocity over time... so.
    [itex]\frac{v_f - v_i}{t} = a[/itex]

    I could get

    vf = vi + at
    this way, but there are still three unknowns! XD...
    I'm sorry , my brain is not in the brightest shape today.
     
  16. Dec 3, 2011 #15

    Doc Al

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    That's the one you need. Hint: It starts from rest.
     
  17. Dec 3, 2011 #16
    Okay, let's see.
    [itex]v_f = v_i + at[/itex]
    I cancel out vi since as you suggested it starts from rest, and put it in a system of equation with the second one.

    [itex]v_f = at[/itex]
    [itex]x_f = x_i + v_i + \frac{1}{2}at^2[/itex]

    And I solve this system of equations, right? Just one thing before I do the math, am I supposed to plug in "6seconds" as time? I don't get why I should do so in the first equation
     
  18. Dec 3, 2011 #17

    Doc Al

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    Some hints:
    In the second equation, your interval is from t = 2 to t = 8 (rewrite it in terms of Δx and Δt).
    In the first equation, use an interval of t = 0 to t = 2, then what you call v_f in that equation equals v_i in the other. (Use the same symbol for both.)
     
  19. Dec 3, 2011 #18
    Okay, I understand, basically the first equation in the very first two seconds give me the initial velocity of the second equation. I am following now.

    So, I can do the following:

    [itex]a= \frac{v_f}{2}[/itex]
    [itex]2000 = 2at + \frac{1}{2}at^2[/itex]

    Where t = 6seconds.

    I get the value of the acceleration to be a = 66,6m/s^2.

    Now I can easily get the initial and final velocity (at t = 2 and t= 8).
    So I could use dv/dt and find the tangential acceleration, then use v^2 / r to get the radial one and find the vector acceleration, right?

    Just one thing I don't get, when I use the radial acceleration formula, what instant am I supposed to plug in for the velocity? Since it has constant acceleration its speed changes within time, so I have to pick a random one, or probably it means at 8s?
     
  20. Dec 3, 2011 #19

    Doc Al

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    Good.

    You've already found the tangential acceleration. (No need for calculus here.)
    Right!

    Reread the problem statement that you posted. They want the values at t = 16 s.
     
  21. Dec 3, 2011 #20
    Right! Thanks a lot Doc AI you have been of great help and I really appreciate it! Finally I understood how to do these kind of problems and once I mastered them I can focus on momentum and such. Thanks a lot!
     
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