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Accelleration Through a Curve

  1. Sep 14, 2006 #1
    If a car gradually changes a direction of 90 degrees at constant speed of 200 m/s over a time period of 20 seconds, then what is the accelleration?

    I don't know if the way I worked it was correct. I think the only assumption is that the acceleration is constant.

    [tex]a_x=\frac{v_xf-v_xi}{t}=\frac{0-200}{20}=-10[/tex]
    [tex]a_y=\frac{v_yf-v_yi}{t}=\frac{200-0}{20}=10[/tex]
    [tex]a=\sqrt{a_x^2+a_y^2}=\sqrt{(-10)^2+10^2}=10\sqrt{2}[/tex]

    Does this seem right to you?
     
  2. jcsd
  3. Sep 14, 2006 #2

    Andrew Mason

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    No. Assume it is a curve of constant radius. Work out the centripetal acceleration. Is there any tangential acceleration here? Take the vector sum of both accelerations.

    AM
     
  4. Sep 14, 2006 #3
    How would someone do that if the radius isn't given? I understand that
    [tex]a_c=\frac{v^2}{r}[/tex]
    but if there is no r value it seems hard to figure out the centripetal acceleration?
     
  5. Sep 14, 2006 #4

    Andrew Mason

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    Use [itex]\theta = \omega t[/itex] and [itex]v = \omega R[/itex]

    AM
     
  6. Sep 14, 2006 #5
    [tex]a_c=\frac{v^2}{r}=\frac{v^2}{v/\omega}=v*\omega=v*\frac{\theta}{t}=200*\frac{\pi/2}{20}=5\pi[/tex]

    Thanks for the help. How come the other way didn't work?
     
  7. Sep 14, 2006 #6

    Andrew Mason

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    Because [itex]\sqrt{2} \ne \pi/2[/itex]. Close but not the same.

    You will notice that the components do not change uniformly with time - it depends on the angle. You are taking the sum of the time average of each component over a quarter turn. Averages over time will be different than the instantaneous value unless the rate of change is uniform over time.

    AM
     
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