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Net displacement of a particle given its equation of motion

  1. Feb 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Hello PF,

    I need some help with the assignment given:
    v=18-2t^2 m/s, where t is in second. When t= 0 the position of the particle is s_0 = - 3 m.
    For the first 5 seconds.
    Determine the total distance ( i got it to 65,33 ft) and the net displacement Δs, and the value of s at the end of the interval.

    3. The attempt at a solution
    I guess I have to use the formula:
    s(t)=s_0 + ∫_0 ^t v(t) dt

    s(t)=3+∫_0^3 18-2t^2 dt
     
    Last edited by a moderator: Feb 12, 2017
  2. jcsd
  3. Feb 12, 2017 #2

    cnh1995

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    That is the correct equation.
    S0= -3m.
     
  4. Feb 12, 2017 #3

    gneill

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    Staff: Mentor

    ... and tf = 5 seconds.
     
  5. Feb 12, 2017 #4

    mfb

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    Staff: Mentor

    Be careful with signs and units and integral borders. Apart from that, the total distance is fine, and the formula for s(t) will help with the net displacement Δs and s at the end.
     
  6. Feb 12, 2017 #5
    But when I integrate it, I do not get the correct answer.
     

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  7. Feb 12, 2017 #6

    cnh1995

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    Why are you multiplying the integral by -3?
    Also, the upper limit of the integral is not 3.
     
  8. Feb 12, 2017 #7
    May bad. So this should be correct?
     

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  9. Feb 12, 2017 #8

    mfb

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    Staff: Mentor

    That number is relevant. Which part did you answer with it?
     
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