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Constant acceleration trajectory

  1. Mar 3, 2015 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    Hello,

    I'v been trying to find an answer for this question for quite some time, I always hit a complex trigonometric equation or a 4th degree polynomial equation. any help would be very much appreciated!
    Here's the problem:

    a point body (B) is thrown in space (no friction) from a point (P1) with an initial velocity vector (V) and a constant acceleration vector (A). The trajectory generated by (B) passes through a point (P2).

    We know the distance between (P1) and (P2) = d
    We know the angle between the vector (P1P2) and the vector (V) = alpha
    We know the magnitude of (V) = v
    We know the magnitude of (A) = a

    Calculate the angle (theta) between (V) and (A)
     
    Last edited by a moderator: Mar 6, 2015
  2. jcsd
  3. Mar 3, 2015 #2

    A.T.

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    Have you tried to decompose v and a into components parallel and perpendicular to P1P2, and then using s = v0i t + 1/2 ai t^2 for each of them? The time t is equal in both equations. while s = 0 for perpendicular and s = d for parallel.
     
    Last edited: Mar 3, 2015
  4. Mar 3, 2015 #3
    Yes, I started by doing that. I ended up with a 4th degree polynomial with long and complex symbolic coefficients. I checked up the general form of the solution for 4th degree equation and it's huge even without long and complex symbolic coefficients. isn't there any easyer way to solve it? Thank you :)
     
  5. Mar 3, 2015 #4
    I think the problem is somehow equivalent to finding the slope of a feild given the slant range and the initial velocity vector of a projectite, but I couldn't go any further from there...
     
  6. Mar 3, 2015 #5

    A.T.

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    An idea, not sure if simpler :

    The trajectory is a parabola (lets assume y-axis is along a). You look for two points on it, with a given distance, and parabola slope at one of them related to the direct slope between them.
     
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