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Another kinematics problem

  1. Sep 30, 2006 #1
    Sorry, I should have posted this in my other thread!

    Here's the information for the question: "A man is running at speed c (much less than the speed of light) to catch a bus already at a stop. At t=0, when he is a distance b from the door to the bus, the bus starts moving with the positive acceleration a.
    Use a coordinate system with x=0 at the door of the stopped bus."

    I have attached a diagram.

    I have been asked a series of related questions from this information, so I will first list the questions previous to the part I am having trouble (Part D) along with my answers to them.

    Part A: "What is [tex]x_{man}(t)[/tex], the position of the man as a function of time?
    Answer symbolically in terms of the variables b, c, and t."

    Using [tex]x = x_{o} + v_{ox}(t-t_{o}) + \frac{1} {2}*a_{x}(t-t_{o})^2[/tex],

    [tex]x_{man}(t) = -b + c(t)[/tex]

    Part B: "What is [tex]x_{bus}(t)[/tex], the position of the bus as a function of time?
    Answer symbolically in terms of a and t."

    Using [tex]x = x_{o} + v_{ox}(t-t_{o}) + \frac{1} {2}*a_{x}(t-t_{o})^2[/tex],

    [tex]x_{bus}(t) = (\frac{1} {2}*a)(t^2)[/tex]

    Part C: "What condition is necessary for the man to catch the bus? Assume he catches it at time [tex]t_{catch}[/tex]."

    [tex]x_{man}(t_{catch}) = x_{bus}(t_{catch})[/tex]

    Part D (having trouble here) : Inserting the formulas you found for and into the condition [tex]x_{man}(t_{catch}) = x_{bus}(t_{catch})[/tex], you obtain the following:

    [tex]-b + ct_{catch} = \frac{1} {2}at_{catch}^2[/tex], or [tex]\frac{1} {2}at_{catch}^2 - ct_{catch} + b = 0[/tex].

    Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man's speed c so that the equation above gives a solution for [tex]t_{catch}[/tex] that is a real positive number.

    Find [tex]c_{min}[/tex], the minimum value of c for which the man will catch the bus.

    Express the minimum value for the man's speed in terms of a and b."

    So I'm really lost on this one. The first thing I noticed was that [tex]\frac{1} {2}at_{catch}^2 - ct_{catch} + b = 0[/tex] is a quadratic, but I'm not sure how to use the quadratic formula to find the minimum speed that the man has to run at. Otherwise, I really don't know how to find his minimum speed using the acceleration of the bus and the distance between the man and the bus at rest. Obviously, the man's speed has to be fast enough so that the bus does not accelerate to an equal or faster speed before he reaches the door. But I'm not sure how to express this mathematically.

    Any hints?
     
  2. jcsd
  3. Sep 30, 2006 #2
  4. Sep 30, 2006 #3
    Nevermind, I figured it out. Just needed to give it a little more thought!
     
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