1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the distance a particle has moved if I know the work done

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    I was asked to find the distance a particle had traveled and was given its integral solved, except the upper limit was an equation.

    [itex](x^2+2x)[/itex] is the force in pounds that acts on the particle


    2. Relevant equations
    [itex]\int_{0}^{x^2+2}(x^2+2x)dx = 108[/itex]


    3. The attempt at a solution
    I know I must find what number the top limit equation equals, but I've been inserting random numbers with no success. Also, I don't know if it is possible to find the value of x and plug it in. This is ridiculous, why couldn't they just give me the integral with the limits and I solve it. :grumpy:
     
  2. jcsd
  3. Feb 26, 2012 #2
    What is the integral of x2 + 2x?
     
  4. Feb 26, 2012 #3
    Its [itex]\frac{1}{3}x^3+x^2[/itex]
     
  5. Feb 26, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Maybe the problem is poor notation; this problem's notation is terrible--it double-uses the symbol x. Much better: write [tex] \int_{0}^{x^2+2}(x'^2+2x')dx' = 108,[/tex] which makes it clear what is the integration variable (x') and what is the unknown (x).

    RGV
     
  6. Feb 26, 2012 #5
    I see what you did there, but how does that makes it easier to solve? I still don't know what to do. :frown:

    Rewinding everything I know:

    The force exerted on the particle: [tex](x'^2+2x')[/tex]

    The initial point of the particle: 0

    The distance traveled by the particle (that's my upper limit) given by: [tex](x^2+2)[/tex]

    And the total work done by the particle after traveling the distance I'm looking for: 108

    Still not helps writing all that. How can I find the upper limit value or the x value to plug it in and get 108 as answer?
     
    Last edited: Feb 26, 2012
  7. Feb 26, 2012 #6
    I agree with RGV. It may be easier to think about using a different variable name for the integral than the x you're trying to solve for.

    Try plugging in the limits of integration as if you are evaluating the integral.
     
  8. Feb 26, 2012 #7
    I kept trying putting in random values and found that the upper value must be 6. In other words:
    [tex] \int_{0}^{6}(x'^2+2x')dx' = 108,[/tex]
    However, isn't there a process to reach that? I did it by brute force.
     
  9. Feb 26, 2012 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You have already evaluated the indefinite integral [itex] F(x') = \int(x'^2 + 2x')\, dx'.[/itex] Using your formula for F, what would be [itex] \int_0^a (x'^2 + 2x') \,dx'[/itex]? What is preventing you from substituting in [itex] a = x^2 + 2[/itex]? Mind you, the final result will not be pretty, but it gives you a place to start.

    RGV
     
  10. Feb 26, 2012 #9
    Here's what I did, perhaps you may find the process of some use.

    1) I computed the integral and applied the limits.
    2) I know that the integral with the limit substitutions = 108
    3) Now it's mostly a matter of expanding the terms and factoring to find solutions.
    4) The only applicable solution I found was x=2
    5) The upper limit with the x=2 substitution is indeed equal to 6.

    Sorry for the short reply, I did it by hand and did the expansions and factoring by calculator since they were quite long. If you would like, I could write it out or post a picture of my solution if you did not understand my steps above.

    Hope that helps,

    Drood
     
  11. Feb 26, 2012 #10
    It gives me x = 2, but I wonder if there is a rule or something for me to find that the value was 6 instead of doing it by guessing like I did.

    Yes please, an image would help a lot. Thanks. Its just that its weird to have to solve it by guessing, but you seem to have some process there.
     
    Last edited: Feb 26, 2012
  12. Feb 26, 2012 #11
    I hope you can read what I wrote. The site kills photo quality apparently and I didn't have enough time to copy it with my scanner.


    IMAG0630.jpg

    Hope that helps,

    Drood
     
  13. Feb 26, 2012 #12
    Oh, I understand what you did there. Thank you very much. I feel dumb now because you make it look so easy and it took me so long.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is the distance a particle has moved if I know the work done
  1. What have I done? (Replies: 1)

Loading...