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rvc

  • one year ago

The acceleration of the train starting from the rest is a= 8/(v^2+1). Find a. v=? when x= 20m b. x=? when v=64.8km/hr

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  1. Michele_Laino
    • one year ago
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    question a) the velocity is given by the subsequent formula: \[\Large v = at\]

  2. rvc
    • one year ago
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    is it the differential equation?

  3. Michele_Laino
    • one year ago
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    please wait

  4. Michele_Laino
    • one year ago
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    yes! I think that we have to write a differential equation

  5. rvc
    • one year ago
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    yes but im confused which to use

  6. Michele_Laino
    • one year ago
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    by definition of acceleration, we can write this: \[\Large a = \frac{{dv}}{{dt}} = \frac{8}{{{v^2} + 1}}\]

  7. rvc
    • one year ago
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    brb

  8. Michele_Laino
    • one year ago
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    the space x(t) traveled is given by this differential equation: \[\Large \frac{{dx}}{{dt}} = v\left( t \right)\]

  9. Michele_Laino
    • one year ago
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    we have to integrate that ODE, so we have: \[\Large x\left( t \right) = \int {v\left( t \right)dt} \] then we have to know the shape of the function: \[\Large v = v\left( t \right)\]

  10. rvc
    • one year ago
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    the question doesn't tell us about the shape

  11. Michele_Laino
    • one year ago
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    yes I know, we have to compute it, nevertheless it is not simple, since after a simple integration, we get: \[\Large \frac{{{v^3}}}{3} + v = 8t\] as you can easily check

  12. Michele_Laino
    • one year ago
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    sincerely I don't know how to overcome this difficulty

  13. rvc
    • one year ago
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    we don't have t in the equation

  14. Michele_Laino
    • one year ago
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    please wait I ask to another helper

  15. Michele_Laino
    • one year ago
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    @mathstudent55 please help

  16. Michele_Laino
    • one year ago
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    @e.mccormick please help

  17. rvc
    • one year ago
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    I think we have these formulas: \[\Large \rm a=v \frac{ dv }{ dx }\] \[\Large \rm v=\frac{ dx }{ dt }\] \[\Large \rm a=\frac{ dv }{ dt }\]

  18. Michele_Laino
    • one year ago
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    yes! nice idea!

  19. Haseeb96
    • one year ago
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    it is very easy I think @Michele_Laino can figure out your this problem. I have to go on my roof. Because there is some huge storm coming.

  20. Michele_Laino
    • one year ago
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    we have: \[\Large \frac{{dx}}{{dv}} = \frac{v}{a} = \frac{{v\left( {{v^2} + 1} \right)}}{8}\]

  21. rvc
    • one year ago
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    Okay @Haseeb96 Thanks bro :)

  22. Michele_Laino
    • one year ago
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    thanks @Haseeb96 actually @rvc helped me with her nice idea

  23. Michele_Laino
    • one year ago
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    ok! we have only to integrate that last ODE, so we can write this: \[\Large x\left( v \right) = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right)\]

  24. rvc
    • one year ago
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    wait how dx/dv? can we write like that way?

  25. Michele_Laino
    • one year ago
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    since: \[\Large \frac{{dx}}{{dv}} = \frac{1}{{\frac{{dv}}{{dx}}}}\]

  26. rvc
    • one year ago
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    oh okay

  27. Michele_Laino
    • one year ago
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    after a substitution, we gave to solve a qudratic equation for v^2

  28. Michele_Laino
    • one year ago
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    we have*

  29. Michele_Laino
    • one year ago
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    for part b) we can use the same formula: \[\Large x\left( v \right) = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right)\]

  30. Michele_Laino
    • one year ago
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    since the motion is the same. We have only to substitute v=64.8 at the right side

  31. rvc
    • one year ago
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    and the first condition?

  32. Michele_Laino
    • one year ago
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    do you mean part a) ?

  33. rvc
    • one year ago
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    yep

  34. Michele_Laino
    • one year ago
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    we have to substitute your data, so we can write this: \[\Large \begin{gathered} 20 = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right) \hfill \\ \hfill \\ 20 = \frac{1}{8}\left( {\frac{{{v^4} + 2{v^2}}}{4}} \right) \hfill \\ \hfill \\ 640 = {v^4} + 2{v^2} \hfill \\ \hfill \\ {v^4} + 2{v^2} - 640 = 0 \hfill \\ \end{gathered} \]

  35. Michele_Laino
    • one year ago
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    now, we can make this change of variable: w=v^2, so we get: \[\Large {w^2} + 2w - 640 = 0\]

  36. rvc
    • one year ago
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    wait 8 X 20 = 160

  37. rvc
    • one year ago
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    oh okayyy

  38. Michele_Laino
    • one year ago
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    now, we have to solve with respect to w, and we have to pick the positive solution of w, since we have to remember that w=v^2

  39. Michele_Laino
    • one year ago
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    I got this: \[\Large w = {v^2} = - 1 + \sqrt {640} \cong - 1 + 25.3 = 24.3\]

  40. Michele_Laino
    • one year ago
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    and hence: \[\Large v = \sqrt {24.3} \cong 5.03\] I write the positive solution only for v, since I believe that v is the magnitude of the velocity of our particle

  41. rvc
    • one year ago
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    i read every step i understand

  42. Michele_Laino
    • one year ago
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    ok! :)

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