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Finding Position with a Quadratic Velocity

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    In the time interval from 0.0 s to 10.8 s, the acceleration of a particle traveling in a straight line is given by ax = (0.1 m/s3)t. Let to the right be the +x direction. The particle initially has a velocity to the right of 10.0 m/s and is located 4.8 m to the left of the origin.


    2. Relevant equations
    (a) Determine the velocity as a function of time during the interval.

    (b) Determine the position as a function of time during the interval.

    (c) Determine the average velocity between t = 0.0 s and 10.8 s.
    3 m/s

    Compare it to the average of the instantaneous velocities at the start and ending times. Are these two averages equal?



    3. The attempt at a solution

    (a) v(t) = (0.05t^2)+10 this was determined to be right...

    (b) x(t) = (0.017t^3)+10t+4.8 this is wrong...why? Significant Figures?
     
  2. jcsd
  3. Sep 17, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    that means for t=0, x= -4.8m (+ve is to the right means -ve is to the left)
     
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