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Seemingly Simple Kinematics Problem II: the Vector Strikes Back

  1. Dec 26, 2006 #1
    Hey, all. A few days ago, I posted a poorly-worded kinematics problem, at https://www.physicsforums.com/showthread.php?p=1195234#post1195234.

    I received a helpful answer practically within minutes, and shortly thereafter realized the answer was irrelevant, and that the fault lay in my inexplicable inability to notice that two of the functions involved could be reduced to simple quadratic equations. So it turns out my question was largely meaningless, and the fault lay with my almost superhuman ability to overlook the painfully obvious.

    Thing is, that fault of mine was an unexpected snag in what I expected to be a downright elementary stage of solving the problem at hand. What comes next is something I'd initially figured would be problematic, and turned out to be worse than I'd anticipated.

    Specifically, I've got to answer the same question with 2- or 3-D vectors, instead of scalars. Where the previous problem had initial velocity, terminal velocity, distance and rate of acceleration, all scalars, this one has an initial velocity vector, a terminal velocity vector, a translation vector representing the terminal position, and an acceleration scalar. Alternately, it has three (assuming we're going for a 3D solution. Lets.) sets of initial speed, terminal speed, and distance, and one acceleration scalar divided between these three axes so that each reaches the desired position and velocity at the same terminal time.

    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 from point a to point b, and be at a certain momentum vector when it reaches point b.

    Given the object's initial velocity vector [tex]v_0[/tex], terminal velocity vector [tex]v_1[/tex], translation vector x, and acceleration rate a, get the object to velocity [tex]v_1[/tex] at point b (as defined by x, with point a as the origin) as quickly as possible.

    In the 1D version of this problem, the object would accelerate until a certain point, then decelerate until it reached the terminus at the desired speed, so solving was simply a matter of finding this point of deceleration. In 2- or 3-D, things are a mite trickier.

    2. Relevant equations

    None come to mind, really. Approaching this from the only angle from which I've a clue, the 1D version of the problem, let [tex]v_2[/tex](given) be the terminal velocity, [tex]v_0[/tex](also given) be the initial velocity, [tex]x_2[/tex](yet another given) be the terminus, and the variables [tex]v_1[/tex], and [tex]x_1[/tex] be the velocity and distance at the point of deceleration, respectively. follows. Unless I made an utterly stupid mistake along the way, the point of deceleration:

    [tex] x_1 = \frac{v_2^2 - v_0^2 + 2ax_2}{4a} [/tex]

    I could demonstrate how I got to that value and those of the variables dependent on [tex]x_1[/tex], but I'm not sure that's relevant to the non-linear problem, and this post's lengthy enough as is.

    3. The attempt at a solution

    None comes to mind. I haven't a clue where to begin expressing an interrelationship this complex, and any attempt to reduce it to simpler component problems fails. Were it not for the variable initial and terminal velocities, it'd be a simple matter of solving for each axis as I did for the 1D problem, comparing the terminal time of each axis to the total time of all three axes, and dividing total acceleration between the axes by this ratio. Unfortunately, I cannot afford to disregard initial and terminal velocities.


    And that's that. Pointing me along a particular line of thought or at some branch or other of mathematics on which to read up would be nearly as helpful as an actual solution, since I imagine I'll be dealing with this sort of problem again in the future. I'm more than willing to learn, I'd just like to avoid going into much more mathematics than I need at this stage. Especially since I'm only certain I know the very basics of integral and differential calculus, and it's been ages since I've properly used those.

    Edit: Incidentally, a solution which can compensate for an additional, constant acceleration vector would be appreciated. By "constant acceleration vector", I don't mean an increase of the acceleration scalar, I mean that the object is constantly being accelerated in a direction which has nothing to do with the trajectory from a to b, and must compensate for this acceleration. Again, pointing me at the branch of mathematics which covers this and giving me some sense of how far I have to go would be nearly as good as a solution.
     
    Last edited: Dec 26, 2006
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  3. Dec 26, 2006 #2

    HallsofIvy

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    Your object accelerates and then decelerates at a constant rate. You don't say that the constant rate is the same for both acceleration and deceleration. I'm going to assume they are a and b respecitively.

    You begin accelerating with constant acceleration a. Your velocity at any time t is at and you distance traveled it (1/2)at2. In particular, your velocity at time t0 is (1/2)at02 and distance from the initial point is . Now you start decelerating from that speed with acceleration -b. You speed at time t is -b(t-t0)+ (1/2)at02 and your distance from the initial point is -(1/2)b(t-t0)2+ (1/2)bt02(t-t0). Set those equal to the given end distance and velocity (momentum/mass) and solve for a, b, t0[/sub]. Of course, for given a, b, there may be no solution or there may be many solutions.
     
  4. Dec 26, 2006 #3
    Sorry, I neglected to clarify that in the 2- or 3-D version of the problem. Yes, the rates of acceleration and deceleration are one and the same.

    I might not be following, but you seem to have misread my rambling and poorly-summarized problem statement. Your velocity at any time until the point of deceleration is [tex]v_0 + at[/tex], and, if [tex]t_1[/tex] is the time at the point of deceleration, your velocity at any point thereafter is [tex]v_0 + at_1 - a(t - t_1)[/tex]. That's in the linear version of the problem, which isn't that hard to solve, even if the answer is that there is no solution. What's actually problematic is trying to do the same in 3D, with a different initial velocity, terminal velocity, and distance for each axis, and the same acceleration/deceleration scalar split between the three in whichever way is optimal. I'm not even sure the point of deceleration would be at the same t for each axis.
     
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