Results 1 to 9 of 9
Like Tree4Thanks
  • 1 Post By skeeter
  • 1 Post By HallsofIvy
  • 1 Post By HallsofIvy
  • 1 Post By HallsofIvy

Thread: Velocity and trains..question a was no problem ...

  1. #1
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    80

    Velocity and trains..question a was no problem ...

    The velocity of a train is given by v = 250 −1/2t^2 .
    (a) What is the displacement of thetrain after 5 hours? and 10 hours?

    I integrated and got the correct answers for both being
    d5 =7375 /6,
    d10 =7000 /3
    (b) At what times is the displacement of the train zero?

    I made 250-1/2t^2 = 0 to find the displacement of the train at 0 but I couldn't work it out...hmm
    maybe my thinking is messed up.

    (c) Describe what’s happening to the train.

    hmmmm....

    Thomas the tank engine...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    15,956
    Thanks
    3577

    Re: Velocity and trains..question a was no problem ...

    $\displaystyle \int_0^T 250-\dfrac{t^2}{2} \, dt = 0$

    $\bigg[250t-\dfrac{t^3}{6}\bigg]_0^T = 0$

    $250T - \dfrac{T^3}{6} = 0$

    Solve for $T$, the time(s) when the train's displacement is zero.

    Do you understand what zero displacement means?
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,109
    Thanks
    2802

    Re: Velocity and trains..question a was no problem ...

    The graph of v= 250- (1/2)t^2 is parabola opening downward with vertex at t= 0, v= 250. That means the train starts with its maximum velocity, 250, then slows down. It will stop (velocity 0) when [tex]t= 50\sqrt{2}[tex] and then start going backwards, with negative velocity. It will be back at its starting point, displacement 0, when 250t- \frac{t^3}{6}= 0, as skeeter says. We can write that as 1500t- t^3= t(1500- t^2). That has three roots, t= 0, when it started, a negative root, which is irrelevant, and a positive root when it will be back at its starting point.
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    80

    Re: Velocity and trains..question a was no problem ...

    Thanks Skeeter I never thought of putting T on the b -a part of the integral...I'll have to keep my eye out for that type of situation..i think 0 displacement means the train hasnt moved from its original position right ?....thanks again it really does make me more aware that I can use any value with letters , numbers and symbols for the integral part something I wasn't aware of ie T to 0
    Cheers !!
    Last edited by bee77; Jul 15th 2017 at 05:49 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    80

    Re: Velocity and trains..question a was no problem ...

    Thanks HallsofIvy The point where its stops comes out at [tex]t= 50\sqrt{2}[tex] so I 'm nt quite sure what that means with the [tex] parts ....but the rest is clear ....where it stops how did you work that out ....thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,109
    Thanks
    2802

    Re: Velocity and trains..question a was no problem ...

    I forgot the "\" in the Latex ending "[\tex]". It should have been t= 50\sqrt{2}.
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    80

    Re: Velocity and trains..question a was no problem ...

    Thanks !! how did you work that out to" stop" the train cheers?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,109
    Thanks
    2802

    Re: Velocity and trains..question a was no problem ...

    I didn't! I made a foolish error and got the wrong answer! Thanks for pointing that out! 250- (1/2)t^2= 0 reduces to 2(250)= 500= t^2 so that t= \pm\sqrt{500}= \pm\sqrt{5(100)}= \pm 10\sqrt{5}, not 50\sqrt{2}! Of course, only the positive answer makes sense in this problem.
    Thanks from bee77
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Mar 2017
    From
    Annabay
    Posts
    80

    Re: Velocity and trains..question a was no problem ...

    No probs , I would've believed it as your math ability is alot higher than mine ...cheers
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. When will the two trains meet.
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: Jun 22nd 2012, 08:47 AM
  2. The two trains and bird distance problem
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Sep 19th 2011, 02:51 AM
  3. Trains
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Dec 15th 2010, 08:37 AM
  4. Replies: 3
    Last Post: Nov 21st 2009, 01:25 PM
  5. Two Trains
    Posted in the Math Challenge Problems Forum
    Replies: 1
    Last Post: Jan 24th 2009, 10:34 AM

/mathhelpforum @mathhelpforum