Yes, that is correct. A man chasing a bus with a constant speed.

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In summary, to catch the bus at a stop, the man must run much less than the speed of light. At t=0, when he is a distance b from the door to the bus, the bus starts moving with the positive acceleration a. xman(t) and xbus(t) are the positions of the man and bus at different times, respectively. Xbus(t) is greater than xman(t) because the bus starts at a different position. The man's speed is constrained to be less than cmin so that he can catch the bus.
  • #1
CellCoree
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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
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
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
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
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
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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