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Homework Help: Trajectory and acceleration.

  1. Nov 7, 2012 #1

    Granted that a body begins accelerating from rest with a constant acceleration a, and after a certain time T changes the direction of its acceleration and begins moving with acceleration -a, would it be correct to say that the body would return to its origin at time t=2T? I mean, are the displacement vectors with respect to time merely 1/2(ax,ay,az)T^2 and -1/2(ax,ay,az)t^2?
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  3. Nov 7, 2012 #2


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    Say you accelerate at 10 for 1 second - your speed would be 10.
    If you then accelerate at -10 for 1 second, you speed would be zero. The distance from the start would still be increasing during this second stage since the velocity is still positive.
  4. Nov 7, 2012 #3
    at time t = T the body will have some velocity v = aT, whereas at time t = 0, the body has a velocity = 0

    so if the body switches instantly from a to -a at time t = T, then it will first have to reduce its velocity to zero before it can start moving in the negative direction

    so no, it will not return to its origin at t = 2T
  5. Nov 7, 2012 #4


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    I think you are alluding here to a formula similar to

    s = ut + 1/2 at2

    For the first time interval "ut" is 0, so your 1/2(ax,ay,az)T^2 is appropriate
    For the second time interval, "ut" is certainly NOT zero, so just -1/2(ax,ay,az)t^2 is insufficient.
  6. Nov 7, 2012 #5
    Would it then be correct to formulate its displacement vector thus:
    r(t) = 1/2(ax,ay,az)T^2 + (-ax,-ay,-az)Tt + 1/2(-ax,-ay,-az)t^2?
  7. Nov 7, 2012 #6


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    I don't agree with the negative signs in the red section.

    The initial velocity for the second segment is the final velocity for the first segment, and in the first segment the accelerations were positive.
  8. Nov 7, 2012 #7
    Is its initial velocity in the second segment not in the opposite direction, as its new acceleration?
  9. Nov 7, 2012 #8


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    yes, but you have negative signs on both the initial velocity AND the acceleration. IF they are in opposite directions, one of them has to be positive [and the acceleration definitely has the negative signs in your nomenclature].
  10. Nov 7, 2012 #9
    Okay, but then I get that its displacement would be equal to zero at t= (1+sqrt(2))*T.
    Could it be? How come the solution for r0, i.e. when body is at the origin, does not include t0=0?
  11. Nov 7, 2012 #10
    I mean, what would be the formulation of its trajectory vector, if in the first segment it is 1/2at^2, whereas in the second segment it is 1/2aT^2+aTt-1/2at^2?
    Is there any way to combine them into one vector?
  12. Nov 7, 2012 #11


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    This problem could mirror the following:

    A rocket is launched and has an upward acceleration of 9.8 ms-2 while the engine is firing.
    Once the engine runs out of fuel, the rocket is (naturally) subject to a downward acceleration of 9.8 ms-2.

    If we were to derive an equation for the displacement, it would be a complex function where you follow one rule for certain domains, and a different rule for other domains.

    0 --> T follow 4.9t2
    T --> X follow 4.9T2 + 9.8T - 4.9t2

    X is the time the rocket hits the ground again.

    Next post will be a conceptual analysis of this.
  13. Nov 7, 2012 #12


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    Conceptual approach to rocket launch.

    While the rocket motor is burning, the speed increases to a maximum (V). the distance covered (height gained) equals the average speed multiplied by the time interval (V/2 x t)
    Once the rocket engine fails, the rocket slows. With the same numeric value of acceleration it will take the same time to stop. Its initial and final speeds will be the same as when it was accelerating up [just V initial and 0 final, rather than 0 initial and V final) so the extra distance travelled up will be an equal amount.

    If the rocket gained height H with the engine burning, it will eventually reach maximum height of 2H

    The s = ut + 1/at2 formula shows us that when u is zero, and s is doubled, t is increased by a factor of √2.
    When this rocket, having reached a height of 2H, falls back to Earth this is indeed the situation.
    So the Rocket will have traveled up, with engine firing, for time T.
    It will have continued to drift up for an additional time T.
    It will then fall back to Earth in time √2.T

    Total trip: (2 + √2)T

    or as you found (1 + √2)T from the end of the first segment of the motion.
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