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Homework Help: Acceleration as a function of distance

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle is travelling along a circular curve with radius 20m. It has an initial speed of 22m/s and then begins to decrease its speed at a rate of a=(-0.25s) m/s2. Determine the magnitude of the acceleration of the particle two seconds later.

    2. Relevant equations

    x2 + y2 = R2

    3. The attempt at a solution
    I integrated the acceleration with respect to s and then set that equal to ds/dt.
    ds/dt = -0.125s2
    Then I moved things around so I had ds/s2 = -0.125dt and integrated both sides. I then end up with a final solution of s=8/t, which i took the derivative of to get velocity, v = -8/t2 and took the derivative again to get acceleration, a = 16/t3. Then I took the original velocity 22m/s and subtracted the velocity found using the above equation, and used this final velocity to find the centripetal acceleration. Then I used the above equation for acceleration to find the component of the acceleration perpendicular to the radius, squared this and added it to the square of the centripetal acceleration, took the square root and well... got the wrong answer. Help?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 11, 2010 #2
    Although I can't quite understand what you have done , I would suggest a much simpler approach wherein you analyse along two perpendicular axes, one perpendicular to velocity and one along it.
    There is constant tangential acceleration along the the direction of velocity. The acceleration along the perpendicular to velocity is variable and a function of the tangential velocity, and hence time as the tangential velocity itself varies with time.
    You basically find out velocity after the given time period, which will depend only on tangential acceleration. This can be used to find out centripetal acceleration. Tangential acceleration is constant. You can add the two vectors.
  4. Mar 11, 2010 #3
    Thats what I was trying to do (find out velocity at that given time), but with the tangential acceleration given as a function of the distance and not the time, I can integrate it to get what the velocity function would be but that is still as a function of distance, not time. So I need to somehow get the velocity as a function of time instead of distance.
    And the tangential acceleration is not constant, a = -0.25s where s is the distance
  5. Mar 11, 2010 #4
    Very sorry ! I missed the 's'.
  6. Mar 11, 2010 #5
    You can differentiate the eqn. a = -0.25*s and obtain the velocity gradient dv/ds. You know a = v* dv/ds and a = dv/dt. A little arrangement and integration will give velocity as function of time and then integrating again you have distance as function of time.
  7. Mar 11, 2010 #6
    This is simply crazy. Rather I am crazy.
    The earlier post was completely wrong. Sorry!
    Actually a = -0.25*s
    a = v*dv/ds.
    Now you see that if you combine the two eqns. you have just v and x in the eqn. v = dx/dt so now there is just x and t in the equation.
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