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Circular Motion of a car

  1. Feb 21, 2006 #1
    A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 237 m, and the car completes the turn in 36.8 s.

    What I've got so far is that I should find the tangential acceleration at B, and then I can use sine and cosine to find the part of the acceleration in each direction, but I can't find the tangential acceleration.... any tips?


    Attached Files:

  2. jcsd
  3. Feb 21, 2006 #2
    If it's travelling at a constant angular speed, how would it have a tangential acceleration??
    And what's being asked?
  4. Feb 21, 2006 #3
    Oh yeah, it's a three part problem... apparently I forgot to put which part I need help with.

    First of all, what is the acceleration when the car is at B located at an angle of 36.8? Express your answer in terms of the unit vectors i and j.

    But ok, I can see how I don't need tangential acceleration since it's constant velocity. But I still need to find centripetal acceleration, which requires velocity.... so how would I find velocity at B?
  5. Feb 21, 2006 #4
    Yup. And just to remind, acceleration is inwards, perpendicuar to v, with the magnitued [tex]mv^2/r[/tex]
  6. Feb 21, 2006 #5
    Right, but I don't have mass or velocity....
  7. Feb 21, 2006 #6
    Whoops, sorry, I've written the force rather than acceleration. Really sorry! It's [tex]v^2/r[/tex]. Why don't you have speed? You were given length of the path and the time for travel. You can get it out of it.
  8. Feb 21, 2006 #7
    I was given time from A to C. This is just measuring from A to B. It's not just half the time given, though, since it's not a 50 degree angle.
  9. Feb 21, 2006 #8
    The speed is said to be constant in magnitude, you can simply divide the total path and time. I can't see the picture since it's still waiting for approval, so I don't know of a 50 degree, but I guess you should be able to exract velocity out of that picture you've attached.
  10. Feb 21, 2006 #9
    Oh it does say the speed is uniform, so I guess I can find the average speed and that would be the speed at B... I'll try that and see if it's right.
  11. Feb 21, 2006 #10
    Actually, in that case, speaking magnitude-wise, average equals to instantaneus. Since rhe angle between path and velocity is always same, we can indeed speak magnitude-wise!
  12. Feb 21, 2006 #11
    Now it says to determine its average acceleration during the 36.8-s interval. The j part of it was the same, but it's telling me the i part of the average acceleration isn't the same, and I can't understand that.
  13. Feb 21, 2006 #12
    You can draw the velocity vector for t=0 and t=36.8, subract them, and divide it to 36.8. And if it's a complete circle, I find it to be 0.
    Did I get you wrong?
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