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Time-dependent Force problem

  1. Feb 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A time-dependent force, Fvecbold.gif = (9.60 ihatbold.gif − 4.10t jhatbold.gif ), where Fvecbold.gif is in newtons and t is in seconds, is exerted on a 2.00-kg object initially at rest. (a) At what time will the object be moving with a speed of 15.0 m/s?

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    I solved for a then integrated that and got 4.8t-1.025t^2

    Set it equal 15 squared both sides to eliminate the square root when solving for the magnitude and finally got 23.04t^2-1.05t^4 -225=0 and can't solve it. So I must have made some mistake.

    [ URLs fixed by a mentor ]
     
    Last edited by a moderator: Apr 19, 2017
  2. jcsd
  3. Feb 21, 2015 #2

    NascentOxygen

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    Staff: Mentor

    Loststudent22, you didn't have the correct URLs for the i and j characters at your webassign site. I fixed them.

    Next time you can try latex. ##{\hat{i}}\ {\hat{j}}##
     
    Last edited: Feb 21, 2015
  4. Feb 21, 2015 #3
    They were the I and J hat vector components. I will have to look into latex I"m not familiar with it yet. I'll trying using a picture next time for the question and my work.
     
  5. Feb 21, 2015 #4

    gneill

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    Staff: Mentor

    Have you tried a substitution for t? You should be able to turn the equation into a quadratic...
     
  6. Feb 21, 2015 #5
    I plugged the equation into wolfram to solve for t and it didn't give a answer
     
  7. Feb 21, 2015 #6

    gneill

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    Staff: Mentor

    You have:

    23.04t^2-1.05t^4 -225=0

    Let : x = t^2

    Write the equation in terms of x. Can you solve that?
     
  8. Feb 21, 2015 #7
    Yeah it doesn't work it gives an answer with imaginary numbers. I pulled out my 89 and it just says false when I try to use the solve function. Is it possible the online assignment gave me numbers that don't work nicely or I must have made some mistake.
     
  9. Feb 21, 2015 #8

    gneill

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    Staff: Mentor

    So you should revisit how you derived your equation. Lay out the steps so we can take a look.
     
  10. Feb 21, 2015 #9
    F=ma
    a=4.8-2.05t=dv/dt
    To arrive at an equation for the instantaneous velocity of object I integrate the above equation

    V=4.8t +c1-1.025t^2+c2

    The object is at rest when t = 0 s. so c1=c2=0

    When V=15 m/s

    The magnitude of V=15=√(4.8t)^2+(1.025t^2)^2

    I square both sides to eliminate the square root and got the equation I'm having trouble with
     
  11. Feb 21, 2015 #10

    gneill

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    Staff: Mentor

    Really? Because I spot a sign difference right away.
     
  12. Feb 21, 2015 #11
    Haha woops I see now + instead of -

    Thanks
     
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