Question

A man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the constant rate of 45 mph . The man can run at a constant rate in either direction to get off the bridge just in time before the train hits him. How fast can the man run?

Answer 1

Answer: 9 mph

Explanation:

Given that a man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the constant rate of 45 mph .

If the man tend to run in the forward direction, he will cover another 2/5 before the train reaches his initial position. The distance covered by the man will be 2/5 + 2/5 = 4/5

The remaining distance = 1 - 4/5 = 1/5

If the man can run at a constant rate in either direction to get off the bridge just in time before the train hits him, the time it will take the man will be

Speed = distance/time

Time = 1/5d ÷ speed

The time it will take the train to cover the entire distance d will be

Time = d ÷ 45

Equate the two time

1/5d ÷ speed = d ÷ 45

Speed = d/5 × 45/d

Speed = 9 mph

Answer 2

The Man needs to run at 9 mph

Explanation:

Let M stand for the man's speed in mph. When the man

runs toward point A, the relative speed of the train with respect

to the man is the train's speed plus the man's speed (45 + M).

When he runs toward point B, the relative speed of the train is the

train's speed minus the man's speed (45 - M).

When he runs toward the train the distance he covers is 2 units.

When he runs in the direction of the train the distance he covers

is 3 units. We can now write that the ratio of the relative speed

of the train when he is running toward point A to the relative speed

of the train when he is running toward point B, is equal to the

inverse ratio of the two distance units or

(45 + M) 3

----------- = ---

(45 - M) 2

90+2 M=135-3 M

⇒5 M = 45

⇒ M = 9 mph

The Man needs to run at 9 mph