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Train, Distance, and SpeedDate: 09/20/97 at 12:01:04 From: jenny Subject: Word problem, finding distances and speeds Bobo is 1/3 of the way across a bridge when he hears a train whistle behind him. A huge locomotive and tons of boxcars are coming at him at 45 mph. Bobo knows that he can make it to the far edge of the bridge at the exact same instant as the train, but he also knows that he can run toward the train and reach the near end of the bridge just as the train gets there. How fast does Bobo run?
Date: 09/20/97 at 12:50:44
From: Doctor Anthony
Subject: Re: Word problem, finding distances and speeds
Train ->45mph A u<- Bobo ->u B
|----------------||-----------|------------------------||
x (1/3)L (2/3)L
You must introduce letters to represent unknown quantities so you
can write down equations connecting times, distances, speeds etc.
The time for Bobo to reach end B of the bridge is (2/3)L/u
The time for the train to reach the same point B is (L+x)/45,
so we can equate these
(2/3)L L + x
------ = ----- ........(1)
u 45
Similarly, the time for Bobo reach A is (1/3)L/u and this equals the
time for the train to reach point A, x/45. Again equating these we get
(1/3)L x
------ = --- ..........(2)
u 45
Now subtract equation(2) from equation(1)
(2/3)L-(1/3)L L + x - x
------------- = ----------
u 45
L L
---- = ----
3u 45
and so 3u = 45 and u = 15 mph
Bobo can run at 15 mph.
An alternative and quicker method is to notice that Bobo has twice the
time to run to point B as he has to run to point A. So the relative
speed of Bobo and the train when Bobo runs towards the train is twice
the relative speed as when he runs away from the train.
This means 45 + u = 2(45 - u)
45 + u = 90 - 2u
3u = 45
u = 15 as before.
-Doctor Anthony, The Math Forum
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