Man catching a bus

  • Thread starter CellCoree
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  • #1
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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

so....

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

is that correct?
 

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  • #2
jamesrc
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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]
 
  • #3
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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?
 
  • #4
jamesrc
Science Advisor
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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)

so:

[tex] X_{\rm bus}(t) = \frac 1 2 at^2 [/tex]
 
  • #5
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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
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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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