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Homework Help: Seemingly simple kinematics problem

  1. Dec 21, 2006 #1
    Not actually homework, but almost definitely not graduate-level stuff. No clue how to use LaTeX, so I'll have to give variables nonstandard names to make this legible.

    1. The problem statement, all variables and given/known data

    An object (say, a spaceship, or something. Something that can accelerate/decelerate at a constant rate) has to travel a certain distance, and be at a certain velocity when it does so. The fastest way for the object to do so would be to accelerate until point h, then decelerate until it's reached the desired distance at the desired speed.

    Given the object's initial velocity i, acceleration/deceleration rate a, target distance x, and target terminal velocity w, find the distance to point h.

    Counterintuitive variable names follow: let y be the time at point h, u be the velocity at point h, and (y + t) be the time at point x

    2. Relevant equations

    h = ((i + u)/2)y = iy + (ay^2)/2
    x = h + ((u + w)/2)t = h + ut - (at^2)/2

    3. The attempt at a solution

    ...is clearly missing something elementary. And will look very, very ugly when expressed without LaTeX. My intentions were straightforward enough: since all the variables are interdependent, reduce them to expressions of the same variable (I opted for u), then express that variable through the given values (i, a, x, and w). Easier said than done.

    ((i + u)/2)y = iy + (ay^2)/2
    (i + u)/2 = i + ay/2
    (i + u)/2 -i = ay/2
    (2((i + u)/2 - i))/a = y
    y = (u - i)/a

    h = ((i + u)/2)y
    h = ((i + u)/2)((u - i)/a)
    h = (u^2 - i^2)/2a

    x = h + ((u + w)/2)t
    x = (u^2 - i^2)/2a + ((u + w)/2)t
    x + (i^2 - u^2)/2a = ((u + w)/2)t
    t = (2x + (i^2 - u^2)/a)/(u + w)

    At this stage, things look a mite messy even in legible notation, but at least we've managed to express each and every variable through u. Now, to expand said expressions and see how they relate to the constants, thereby getting the powerup and winning the game.

    x = h + ut - (at^2)/2
    x = (u^2 - i^2)/2a + u(2x + (i^2 - u^2)/a)/(u + w) - 0.5a((2x + (i^2 - u^2)/a)(2x + (i^2 - u^2)/a)/((u + w)(u + w)))

    Oh. Oh, my. At this stage, I have a sneaking suspicion the expression at the end of all this will be trivial, and therefore not what I'm looking for at all. And yet the expansion of t^2 alone is

    (4x^2 + (4xi^2 - 4xu^2)/a + (i^4 - 2u^2i^2 + u^4)/a^2)/(u^2 + 2wu + w^2)

    I haven't a clue how to begin extracting u from that mess. I may be relying on intuition overmuch, but I can't see how a problem this straightforward became this complex in my hands. Any advice would be appreciated.
  2. jcsd
  3. Dec 21, 2006 #2


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    Staff: Mentor

    Welcome to the PF, The 'Hoff. Thanks for using the Homework Posting Template.

    First, I learned LaTex very quickly and easily by reading the start of the sticky post about it at the top of our Tutorials forum:


    I printed out a copy of one of the crib sheets, and keep it tacked on the wall of my office.

    Second, for my initial help on this, I'd suggest making a simple drawing of the velocity versus time to see if that helps your intuitive reasoning. You know that the velocity at the end has to equal zero, and that the slope of the changing velocity plot has to be the same for the deceleration as for the acceleration. So work backwards from the final time value, draw an upslope from right-to-left, then a matching downslope from right-to-left, finishing at the height of the initial velocity. That plot should give you some intuition for what happens as the initial velocity is increased and decreased, and as the allowed constant accel/decel changes. Does that make it easier to get the math to work out?
  4. Dec 21, 2006 #3
    'preciated. I really should have just taken a few minutes to look around, but I was in too much of a rush to do more than run a quick search and see whether or not my sorta question had already been posted.

    On it. Or at least, plan to be on it shortly. Busy night. Thanks for the tip.
  5. Dec 21, 2006 #4


    User Avatar

    Staff: Mentor

    BTW, I just saw this cool tip for searching the PF. The PF search engine is pretty basic -- you can't search on phrases or multiple keywords AND'ed together. If you enter more than one keyword, it just OR's them, so you get lots of unhelpful hits. You really need a unique keyword to use the standard PF search engine.

    But I saw this great tip someplace in the forums earlier today, about using a google feature to search the PF forums. Keep it in mind for future PF searches! :approve:

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