```Question 166738
a pedestrian is three eights of the way across the a train bridge when he hears a train coming! If he runs as fast as possible back toward the train, he will get off just in in time to avoid a collision. Also, if he runs as fast as possible away from the train, he will get off the bridge (on the other side) just in time to avoid a collision.
If the train is traveling at 60 miles per hour. How fast does the person run?
:
:
Train----------d------------Br.....3x......P..........5x.........Br
:
Let d = dist train is from the bridge
Let 8x = length of the bridge
Pedestrian is 3x from one end and 5x from the other end of the bridge
:
Let s = running speed of p (mph)
:
Toward the train time equation (arrive at end of the bridge at the same time)
{{{d/60}}} = {{{(3x)/s}}}
:
Away from the train time equation (arrive at other end at the same time)
{{{((d+8x))/60}}} = {{{(5x)/s}}}
:
After much manipulation of these equations (too much, to record here) I came up with:
d = 7.2
8x = 4.8 mi
P = 1.8 mi from one end and 3 mi from the other end
:
s = 15 mph running speed
:
:
Check this using the "away from the train" situation:
{{{((7.2+4.8))/60}}} = {{{3/15}}}
{{{12/60}}} = {{{3/15}}} = .2 hrs
:
Check using the "toward the train" situation
{{{7.2/60}}} = {{{1.8/15}}} = .12 hrs

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