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Far out free response concerning motion

  1. Jan 17, 2006 #1
    Hello everyone, I have yet another problem that I just can't seem to work my way around. Here it is:

    A particle starts from rest and at the origin. The particle is cofined to motion along a straight line. The particle accelerates uniformly with magnitude a for a time interval t, then travels at a constant velocity for the same time interval t, followed by a constant decceleration of magnitude -a for a time interval T, different from that of t. That's the question. Here's the problem;

    In terms of the given time interval t, at what time does the particle return to the origin?

    Okay, I understand that there are multiple ways of looking at this problem. I approached as a graphical solution with the motion formulas (kinematics if you will). I made a velocity vs time graph, superimposed on the accel vs time graph and represented that the positive trapezoid had an equal area as the triangle below the time axis. the issue lay in the algebra of the kinematics and the adding of the displacements. My final answer was in a strange quadratic form, but I have a feeling that the answer is like 7t....

    Can anyone point me in the right direction?? Thanks a lot

    P.S. Can someone explain to me how to put in the cool math lettering, um I think it's called like latex or something...thanks again
     
  2. jcsd
  3. Jan 17, 2006 #2
    x1 = 1/2at^2
    x2 = vt = att = at^2
    x3 = 1/2 at^2 where x3 is the displacement again in the forward direction with decceleration and at this instant the instantaneous velocity is 0
    thus you have to travel 2at^2 backward with an acceleration a more correctly in the backward direction. If the time to travel 2at^2 is y, then
    T = y + t.
     
  4. Jan 18, 2006 #3

    andrevdh

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    One way of solving for the time is to divide the motion into three seperate parts. For each of these you define an equation that describes the displacement as a function of time. The third part of the motion starts out with a velocity [itex]v[/itex] and the object is decelerated at [itex]-a[/itex]. Setup the equation for this part of the motion from the point where the decelerating force is applied, that is [itex]x=0[/itex] at the beginning of this phase. Then add up the total displacement for the first two parts, [itex]x_{12}[/itex], in terms of [itex]t_1[/itex] and [itex]v[/itex]. Then you need to find the elapsed time for the object's displacement to become [itex]-x_{12}[/itex] during the third part. You should end up with an quadratic equation in [itex]t[/itex] with only one positive root in terms of [itex]t_1,v\ and\ a[/itex].
     
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