Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

ODE trouble

  1. Jan 24, 2005 #1
    I am having trouble with the following problem:

    A rocket sled having an initial speed of 150mi/hr is slowed by a channel of water. Assume that, during the braking process, the acceleration a is given by a(v)=-u*v^2, where v is the velocity and u is constant.

    a) write the equations of motion in terms of v and x

    b) if it requires a distance of 2000 ft to slow the sled to 15 mi/hr, determine the value of u

    c) Find the time required to slow the sled to 15 mi/hr


    I’m having trouble with part c) because I can’t find a way to get time into the experesion.

    In part b I found v(x)=150e^(-u*x) but I don’t know where to go with this to find time

    I would really appreciate some help

    thanks
     
  2. jcsd
  3. Jan 24, 2005 #2
    [tex]t=\int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v(x)}[/tex]
     
    Last edited: Jan 24, 2005
  4. Jan 24, 2005 #3
    thanks alot for the hit. It was what i needed to solve the problem
     
  5. Jan 24, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    You have no reason to thank him for.He gave you the erroneous formula for the solving...

    [tex] \frac{dv(t)}{dt}=-u v^{2}(t) [/tex]

    Can u find the v(t)...??Then integrate and find x(t).

    Daniel.
     
  6. Jan 24, 2005 #5
    This was not the intention of the question. First he had to rewrite the DE in terms of v and x, and then solve for v(x) not v(t). He already did that and now for part c) has to find the time using his results of a and b. And the formula I gave him was not erroneous!
     
  7. Jan 24, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    And how would you determine [itex] v(x) [/itex],without knowing [itex] v(t) [/itex] and [itex] x(t) [/itex] ???????????

    Daniel.
     
  8. Jan 24, 2005 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Please explain the last step of your equation...Why is it [itex]v(x) [/itex] instead of [itex] v(t) [/itex] ...?? :rolleyes:

    Daniel.
     
  9. Jan 24, 2005 #8
    And he succeeded; he got as a result:

    It was all in his post!
     
  10. Jan 24, 2005 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Well,the issue remains...Your advice was wrong...And naturally it couldn't have lead to any VALID result...

    Daniel.
     
  11. Jan 24, 2005 #10
    Because he calculated v(x) and not v(t) you can't use:

    [tex]x(t)=\int v(t) dt[/tex]

    And since he has v(x) he can use the definition of velocity v=dx/dt so dt=dx/v and integrate.
     
  12. Jan 24, 2005 #11
    Why is the formula I gave him not valid?
     
  13. Jan 24, 2005 #12

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Check post number 7.

    Daniel.
     
  14. Jan 24, 2005 #13
    You can easily see the validity of:

    [tex]t=\int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v(x)}[/tex]

    By writing it using a Riemann sum. On every interval dx the amount of time it takes him is inversely proportional to the velocity, wich he obtained a s a function of x. Summing over all intervals dx gives him the total time.

    He already solved the DE to find v(x), so now he can use v(x) instead of v(t).

    Try it your way and see if you get the same result...
     
  15. Jan 24, 2005 #14

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The integrands are:
    [tex] \frac{dx}{\frac{dx(t)}{dt}} =\frac{dx}{v(t)} [/tex]

    Not v(x).

    Daniel.

    P.S.There's no need Riemann sum whatsoever.
    [tex] v(t)=\frac{dx}{dt}[/tex]=>what i've written above,which of course is an enormity...

    P.P.S.The correct integration would be
    [tex] \int dx=\int v(t)dt [/tex]
    which would give u x(t),knowing v(t)...
     
    Last edited: Jan 24, 2005
  16. Jan 24, 2005 #15
    Ofcourse solving the DE for v(t) is also an option, but not the intention of the question. He already solved the the DE for v(x). Now if somebody asked:

    Then the answer is [tex]t=\int_0 ^{x_1} \frac{dx}{v(x)}[/tex]. Or why not???!!
     
    Last edited: Jan 24, 2005
  17. Jan 24, 2005 #16
    Well.....?
     
  18. Jan 24, 2005 #17

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Because it's wrong...Forget about the definite integral.Let's discuss variables in the differentials
    U're saying that
    [tex] dt=\frac{dx}{v(x)} [/tex]
    .Is that right,or am missing something...?? :rolleyes:

    Daniel.
     
  19. Jan 24, 2005 #18
    Well maybe that's mathematically not so nice. But given v(x) wouldn't the expression I advised the OP to use work?!
     
  20. Jan 24, 2005 #19

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    It's not "mathematically nice",because it's mathematically incorrect.
    From what u've considered,by switching fractions,u'd get
    [tex] v(x)=\frac{dx}{dt} [/tex]

    which is incorrect...

    Daniel.

    P.S.So i was right all along,eh...???From my first message i asserted the advice was wrong and hence useless...
     
  21. Jan 24, 2005 #20
    Why isn't this correct? dx/dt is the definition of velocity and he happens to have v as a function of x, not t. And why is the formula I gave him wrong? Is the result he gets for t incorrect? [So is the formula I gave him indeed useless?!]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: ODE trouble
  1. Trouble with an ODE (Replies: 4)

  2. ODE Review (Replies: 3)

  3. The solution of ODE (Replies: 5)

Loading...