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Man catching a bus

  1. Sep 3, 2004 #1
    A man is running at speed [tex]c[/tex] (much less than the speed of light) to catch a bus already at a stop. At [tex]t=0[/tex], when he is a distance [tex]b[/tex] from the door to the bus, the bus starts moving with the positive acceleration [tex]a[/tex].
    Use a coordinate system with [tex]x=0[/tex] at the door of the stopped bus

    i have two questions :frown:

    What is [tex]x[/tex]man[tex](t)[/tex], the position of the man as a function of time? Answer symbolically in terms of the variables [tex]b[/tex], [tex]c[/tex] , and [tex]t[/tex] .

    [tex]x[/tex]man[tex](t)[/tex] = _____________

    (picture of the man and the bus is also included)
    because the man's speed is constant, i used the formula

    x(t) = x(0) + vt


    [tex]x[/tex]man[tex](t)[/tex] =0+ct

    is that correct?

    Attached Files:

  2. jcsd
  3. Sep 3, 2004 #2


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    From the picture, it looks like the position of the man at time = 0 is x = -b. So the answer should be [tex] x_{\rm man}(t) = -b + ct [/tex]
  4. Sep 3, 2004 #3
    your right,thanks.

    i have another question

    what is [tex]X[/tex]bus[tex](t)[/tex] the position of the bus as a function of time?

    using the formula x(t) = x(0) +v(0)t + (1/2)at^2

    x(t) = b +v(0)t +(1/2)ct^2

    hmm what would be v(0)t? and is my setup correct? and what should i do next?
  5. Sep 3, 2004 #4


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    That's the right formula, but think about what each variable means:

    x(0) for the bus is 0, because the bus starts at x = 0.

    v(0) for the bus is zero because it starts from rest

    a for the bus is a (as stated in the problem)


    [tex] X_{\rm bus}(t) = \frac 1 2 at^2 [/tex]
  6. Sep 4, 2004 #5
    Inserting the formulas you found for [tex]x[/tex]man[tex](t)[/tex] and [tex]x[/tex]bus[tex](t)[/tex]into the condition [tex]x[/tex]man[tex](t catch)[/tex] = [tex]x[/tex]bus[tex](t catch)[/tex], you obtain the following:

    [tex]-b + ct[/tex]catch[tex]=1/2at^2[/tex]catch

    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 [tex]c[/tex] so that the equation above gives a solution for [tex]t[/tex]catch that is a real positive number.

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

    how do i find [tex]c[/tex]min?
  7. Sep 4, 2004 #6


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  8. Sep 6, 2004 #7
    so your telling me to slove for cmin^2 -2ab =0 right?

    so cmin = sqrt(2ab). is that the correct answer?
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