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Displacement given position function at time t

  1. Sep 20, 2016 #1
    1. The problem statement, all variables and given/known data

    The position function of a projectile is given by r(t) = (5t + 6t2) m I + (30 - t3) m j. What is the displacement of the particle in magnitude angle form at the = 2.0 seconds.

    2. Relevant equations
    √x2+y2
    tan-1 = y/x
    Possibly: x = v0t + 1/2at2


    3. The attempt at a solution

    I assumed you could calculate the position at t = 2.0 s by plugging into the position function and subtract the initial position. So the initial position would be 30 in the positive you direction (j) and 0 i. The position at t = 2.0 s would be 10 + 6 * 4 = 34 i and 30 - 8 = 22. So the displacement would then be 34 m in the positive x direction and 8 meters in the negative y directions The resultant magnitude would be √(34)2+(-8)2 = √1156 + 64 = √1220.

    The angle would be tan-1θ = y over x then

    Is this correct? I haven't encountered another problem finding displacement given position function. Or should I find initial velocity and acceleration and use x = v0t + 1/2at2.
     
  2. jcsd
  3. Sep 20, 2016 #2

    gneill

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    Staff: Mentor

    Hi Robophys, Welcome to Physics Forums!

    Your method looks good. Take care when rounding results; your value of 34 m for the displacement looks to me like you truncated rather than rounded. Of course, it might also be that you rounded intermediate results along the way and lost accuracy in the calculations. So beware of that! Keep extra "guard digits" in intermediate steps. Only round at the end to display final results.
     
  4. Sep 20, 2016 #3

    kuruman

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    Homework Helper
    Gold Member

    Welcome to PF Robophys.

    Your method is correct. Displacement is the difference between final and initial position.
     
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