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Newton's 2nd Law Problem

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A 0.25 kg particle moves in an xy plane according to x(t) = -15 + 2t - 4t3 and y(t) = 25 + 7t - 9t2, with x and y in meters and t in seconds. Find formulas at time t<35 sec for the (a) the magnitude and (b) the angle (relative to the positive direction of the x axis) of the net force on the particle (in radians), as well as (c) the angle of the particle's direction of travel (in radians)

    2. Relevant equations
    F=ma
    Pythagorean thereom
    Fx=Fcosθ
    Fy=Fsinθ

    3. The attempt at a solution
    With the x(t) and y(t) being the decoupled position vector, I thought I would use Pythagorean Thereom to come up with the vector r
    r=((-4t^3+2t-15)^2+(-9t^2+7t+25)^2)^0.5
    r=(16t^6+65t^4-6t^3-397t^2+290t+850)^.5
    approximating the square roots for the coeffecients:
    r=4t^3+9.06t^2-2.45t^1.5-19.92t+17.03t^0.5+29.15
    I have a strong suspicion that I went about this in a very wrong way
     
  2. jcsd
  3. Sep 16, 2016 #2
    As Force=mass*acceleration and the acceleration in each direction is the second derivative of the decoupled position function the acceleration functions would be ax(t)=-24t ay(t)=-18
     
  4. Sep 16, 2016 #3
    Making the force function=((576t^2+324)^.5)/4 ???
     
  5. Sep 16, 2016 #4

    ehild

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    Homework Helper
    Gold Member

    Yes.
     
  6. Sep 16, 2016 #5
    Okay so the angle of the force would then be .25 arctan(-18/-24t) ??
     
  7. Sep 16, 2016 #6
    or .25 arctan(3/4t) simplified and the angle of the velocity being arctan(-18t/.12t^2)=arctan(3/2t)
     
  8. Sep 16, 2016 #7
    Those last two were incorrect.
     
  9. Sep 16, 2016 #8

    gneill

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    Staff: Mentor

    Pay attention to the quadrant that the acceleration vector lies in. The arctan function can be a trap when the signs of the x and y components are not both positive. You need to sort out the correct quadrant yourself and adjust the result accordingly.
     
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