MATH PROB. - Train Bridge (S447b)
From: William Wu of U. C. Berkeley
A man is
3/8's of the way across a train bridge, when he
I did this very
differently than your formal solution
If I run towards
the train I cover 3/8 the length of the bridge
Thank you Jack.
I could not imagine how you could do this problem
= the width of the bridge
Distance equals Rate times Time ( D = R x T )
Time equals Distance divided by Rate ( T = D/R )
Time 1 of train = time of man returning ( Y / R = 3/8 (X) / 15 )
Time 2 of train = time of man going forward ( (Y + X)/R = 5/8 (X) / 15 )
That's two equations and three unknowns, but it solves.
Cross multiplying yields
15 Y = 3/8 X R and 15 Y + 15 X = 5/8 X R
Substitute the 15 Y from the first equation into the second yields
3/8 X R + 15 X = 5/8 X R
Divide everywhere by X.
3/8 X + 15 = 5/8 X
15 = 1/4 X
60 mph = X
The train is going 60 mph no matter the width of the bridge.
The only reason
I worked the problem was the site host's statement