Work Question Two

Question

A man is running through a train tunnel. When he is 2/5 of the way through, he hears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in mph, is he running? 

Answer 1

Let x be the total length of the tunnel, y be the distance the train is from the entrance of the tunnel, and v the rate this doofus is running at. The train speed is 60 mph and the guy is 0.4x into the tunnel at t=0. Length is measured in miles, time in hours, and speed in mph.

 

1. Time for train to reach tunnel entrance =  time for the guy to run back to the entrance: y/60=0.4x/v

2. Time for train to reach tunnel end = time for the guy to run to end of tunnel: (x+y)/60=0.6x/v

 

From (2) we have v=0.6x*60/(x+y), which simplifies to v=36x/(x+y)

From (1) we have y=0.4x*60/v, which simplifies to y=24x/v

Inserting the latter into the former, we obtain v=36x/(x+24x/v), which simplifies to v=36/(1+24/v).

Multiplying both sides by the denominator, we get v(1+24/v)=36, which equates to v+24=36, and hence v=12 mph.

 

Answer: The guy runs at a rate of 12 mph (pretty fast!).

 

Cheers,

Alex