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Displacement, velocity, acceleration

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    consider a particle moving in the x-y plane as follows

    x = 5t^2 - 3t
    y = 2 + t^2

    a) find the mag of the displacement, vel, and accel at t = 2
    b)find the time when the vel and the accel are at right angles to each other
    c) find the average vel between t = 0 and 1

    2. Relevant equations



    3. The attempt at a solution

    a.... x(2) = 14, y(2) = 6, mag of displacement is sqrt(14^2 + 6^2) = 15.23

    vx(2) = 10(2) - 3 = 17, vy(2) = 2(2) = 4, mad of disp = 17.46

    ax(2) = 10, ay(2) = 2, mag of disp = 10.20

    b.... i tried to graph them on my calc and find the intersection but i couldnt find the intersection... is this an appropriate method

    c..... average veolicy of x from t = 0 to 1 is... (7-(-3))/(1-0) = 10, for y... (2-0)/(1-0) = 2
    the mag of the average is sqrt(10^2 + 2^2) = 10.20 m/s
     
  2. jcsd
  3. Sep 3, 2009 #2

    rl.bhat

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    b) When the velocity acceleration are perpendicular, there is no change in the velocity. Take the derivative of velocity and equate it to zero.
     
  4. Sep 9, 2009 #3
    "b) When the velocity acceleration are perpendicular, there is no change in the velocity. Take the derivative of velocity and equate it to zero. "

    im a bit confused. the derivative of the velocity is a constant. does that mean there is no answer.


    also is part c correct. it is the same answer as the mag of acceleration. is coincidence or what
     
  5. Sep 9, 2009 #4

    rl.bhat

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    Find vx and vy and ax and ay.
    v = ( vx^2 + vy^2)^1/2
    Find dv/dt and equate to zero. Since a is not equal to zero, magnitude of the velocity may be constant and acceleration is perpendicular to v.
     
  6. Sep 9, 2009 #5
    vx = 10t - 3; vy = 2t

    v = [tex]\sqrt{(10t-3)^2 + (2t)^2}[/tex]
    v' = .5(104t^2 - 60t + 9)^(-.5)(208t-60) = 0
    i set just the top equeat to zero, 208t - 60 = 0, t = .29
     
  7. Sep 9, 2009 #6

    rl.bhat

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    Your v' is wrong.
    It should be
    v' = 0.5*{(10t-3)^2 + (2t)^2}^-1/2*{2(10t - 3) + 4t}
    Now equate it to zero. Since the denominator cannot be zero, 2(10t - 3) + 4t = 0
     
  8. Sep 10, 2009 #7
    "Now equate it to zero. Since the denominator cannot be zero, 2(10t - 3) + 4t = 0 "

    if x = (10t-3)2+(2t)2, then dx = 2(10t-3)(10) + 2(2t)(2) = 200t - 60 + 8t = 208t-60
     
  9. Sep 10, 2009 #8

    rl.bhat

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    Here v = Sqrt[ (10t-3)2+(2t)2]
    What is the derivative of this?
     
  10. Sep 12, 2009 #9
    but if u work out all the math inside the sqrt, you get

    sqrt(104t^2 - 60t + 9)
     
  11. Sep 13, 2009 #10
  12. Sep 13, 2009 #11

    rl.bhat

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  13. Sep 13, 2009 #12

    rl.bhat

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    I have edited this post.
     
  14. Sep 13, 2009 #13
    no problem, thanks for all the rest of ur help
     
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