# Man catching a bus

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 two questions What is $$x$$man$$(t)$$, the position of the man as a function of time? Answer symbolically in terms of the variables $$b$$, $$c$$ , and $$t$$ .

$$x$$man$$(t)$$ = _____________

(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....

$$x$$man$$(t)$$ =0+ct

is that correct?

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jamesrc
Gold Member
From the picture, it looks like the position of the man at time = 0 is x = -b. So the answer should be $$x_{\rm man}(t) = -b + ct$$

i have another question

what is $$X$$bus$$(t)$$ 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?

jamesrc
Gold Member
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:

$$X_{\rm bus}(t) = \frac 1 2 at^2$$

Inserting the formulas you found for $$x$$man$$(t)$$ and $$x$$bus$$(t)$$into the condition $$x$$man$$(t catch)$$ = $$x$$bus$$(t catch)$$, you obtain the following:

$$-b + ct$$catch$$=1/2at^2$$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 $$c$$ so that the equation above gives a solution for $$t$$catch that is a real positive number.

Find $$c$$min, the minimum value of for which the man will catch the bus

how do i find $$c$$min?

so your telling me to slove for cmin^2 -2ab =0 right?

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