[superprismatic]

A man is running on a railroad trestle bridge. He has traversed .6 of

the length of the bridge when he spots a train ahead of him coming at

him. If he continues to run toward the train at a constant speed of

15 miles per hour, the train and he will meet at the end of the bridge.

If, however, he instantaneously turns around and goes back from whence

he came at a constant speed of 15 miles per hour, the train and he will

meet at the other end of the bridge. Assuming that the train's speed

is constant, what is the speed of the train?

[Guest]

75 mph

[dark_magician_92]

it is 75kmph,

Edited August 31, 2011 by dark_magician_92

[Guest]

If length of Bridge = B

Distance from train to bridge = D

and Train speed = T (mph)

If he runs toward the train, the collision occurs when he reaches the edge of the bridge = 0.4B/15. This is the same time the train reaches the edge of the bridge = D/T

If he runs away from the train, the collision occurs when he reaches the other edge of the bridge = 0.6B/15. This is the same time the train reaches the far edge ((D+B)/T

So, just solve the system of two equations for T

(1) 0.4B/15 = D/T

(2) 0.6B/15 = (D+B)/T

T=75mph

Edited August 31, 2011 by CrayolaSunset

[Aaryan]

a dire situation for the man

[Guest]

x = length of bridge.

man turns and runs from train. after covering .4x the train reaches the bridge.

Man has .2x left to cover to reach other end.

So train must travel five times as fast as man to meet him at other end.

Therefore, train travels at (5*15)mph = 75mph ( so dark_magician_92 is incorrect! ).

Can anyone see from the above how far the train was from the edge of bridge when man first saw it?

[Guest]

spoiler for answer the the question in my previous post above.

when the man first saw the it the train must have been 2*(length of bridge) from the bridge.