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Homework Help: Maximum Positive Coordinate Reached by a Particle

  1. Dec 29, 2015 #1
    • Missing template due to originally being posted in different forum.
    The problem gives you a function describing the position of a particle moving along an x axis:
    x(t) = 12t2 - 2t3

    With this function, one must determine the maximum positive coordinate reached by a particle and the maximum positive velocity. The first step to the problem is to take the derivative of the original function to obtain
    v(t) = 24t - 6t2

    And then again to obtain
    a(t) = 24 - 12t

    Where x is position, t is time, v is velocity and a is acceleration.

    According to the solution manual for this specific problem, maximum positive coordinate reached by the particle requires v = 0, by which we then solve for t which turns out to be t = 4s. We then plug in the value for t in the original function x(t) and yield xmax = 64m. Finding maximum positive velocity is similar, whereas
    a = 0.

    What I don't understand is why the maximum positive coordinate for x is at v = 0, likewise for maximum velocity at a = 0.

    I understand, mathematically, that this has to do with finding the min and max of a function by finding its critical points. In the function x(t) the term -2t3 is growing faster than 12t2, therefore, eventually, x(t) will become negative.

    At t = 0 a particle following this path is x = 0, v = 0 and a = 24.
    The particle ascends gradually in the x(t) graph.
    At t = 2, a = 0 so v must be constant (v = 24) and x is midway between its starting point and its highest point. In a v(t) graph, this point is the peak (vertex) of a downwards parabola. At this same point, the down-sloping line of an a(t) graph intercepts with the t-axis.
    At t = 4, x(t) reaches the peak of its trajectory. v = 0 so the v(t) graph intercepts the t-axis and a = -24 (shouldn't a = 0 if v = 0?)
    At t = 6, the particle reaches the ground so x = 0 and -just before landing- v = -72, and a = -48

    Of course, we are ignoring factors light the drag force caused by air. When the projectile is in the air, the only force acting upon it is g and so its velocity gradually decreases until it reaches its highest point; therefore, with classical physics, we can conclude that the maximum position of a particle in this trajectory is at v = 0 because at this point, the projectile begins to descend back towards the ground...

    Actually, in the process of writing this I just realized that I interpreted this the wrong way. I'm assuming the particle is a projectile (the shape of the graph confused me as it rises and falls, which gave me said impression), however it is moving on the x axis; than again, this may be fixed by substituting x for y. It is then that I must pose the question, what if the particle is actually a marble rolling on a surface? Then, instead of considering gravity, we must consider friction, correct? With that said, friction would not explain why the marble accelerates in the opposite direction. In that case, going back to gravity and ignoring friction, we must assume that the marble is actually rolling up a hill and then back down.

    To be honest, at this point, I question whether I'm just spouting gibberish. I'm sorry if my ideas seem to be all over the place, I'm trying to portray my thought process in an orderly manner.

    Moving on, I think I've convinced myself why x has reached its highest point at v = 0 whether it represents a vertical or horizontal path. But what if the function x(t) is not describing motion, but is actually a marketing formula where x is a product and t is the price? In this case, we ask ourselves what price yields maximum product sold? I'm sorry, I don't know much about marketing so this entire set up might not make sense. Perhaps my error is trying to adjust this formula to other hypothetical scenarios where it might apply?

    I guess, to make my question less ambiguous, is x maximized at v = 0 regardless of the function or the event under observation? This idea sounds wrong just saying it, I think, but my ideas are so convoluted that I can't think straight anymore. Again, sorry for this mess. I might be asking the wrong question to begin with.
  2. jcsd
  3. Dec 29, 2015 #2


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    "v" changes sign at its max/min.
  4. Dec 30, 2015 #3


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    You get the critical points of a function F(t) where the derivatives dF/dt are zero. You can visualize it as the the tangent to the function. The tangent line is horizontal at the maximum and at the minimum of the function.
    The velocity is defined as time derivative of displacement, v=dx/dt. The acceleration is defined as a the time derivative of the velocity, a=dv/dt. The acceleration is zero at the critical point(s) of the v(t) function, and the velocity is zero at the critical point(s) of the x(t) function.
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