SOLUTION: Could you please help me answer this question: A train traveling at 60 miles per hour takes 5 seconds to enter a tunnel and another 40 seconds to pass completely through the tunnel

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Question 970173: Could you please help me answer this question: A train traveling at 60 miles per hour takes 5 seconds to enter a tunnel and another 40 seconds to pass completely through the tunnel. How long is the train ( in feet) and how long is the tunnel? Thank you so much
Found 2 solutions by josmiceli, Evy Hofbauer:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Picture the tunnel being quite a bit longer than the train
The problem says the train " takes 5 seconds to enter a tunnel "
I think this means that the end of the train reaches the beginning
of the tunnel in 5 seconds
+60+ mi/hr = +60%2F60+=+1+ mi/sec
+d+=+1%2A5+
+d+=+5+ mi
The train is 5 mi long
---------------------
The back of the train reaches the other end of the tunnel
in +40+ sec
The head of the train reached the end of the tunnel +5+
sec before this, or in +35+ sec
+5+ sec later the back of the train reaches
the end of the tunnel
------------------------------
+35+%2B+5+=+40+ sec
+d+=+1%2A40+
+d+=+40+ mi
The tunnel is 40 mi long
-----------------------
Hopd I got it
and hope this makes sense
-----------------





Answer by Evy Hofbauer(7) About Me  (Show Source):
You can put this solution on YOUR website!
in school they probably want you to start with 60 mi/hr × 1 hr/60 min and cross out the alike units each time making a really long fraction and multiplying the tops seperate from the bottom then simplifying.
But a really simple way to deal with fraction conversions is to replace them each with what they are equal to and simplify. It can't be done in 1 step but the units are already done so tell me what you think

60 mph = 60 miles/60 min = 1 mile per minute
1 mile/ 1 min = 5280 feet/ 60 seconds = 88 feet/sec
anyway it takes 5 seconds to enter the tunnel and is passing 88 feet of itself through at each second so 5sec* 88 feet/sec = 440 feet
So almost 10 boxcars which makes sense if anyone is wondering.
the length of the tunnel is...
If you are thinking it is a trick question and the front is already 5 seconds or 440 feet into the tunnel and you have to add that to 40 seconds you are actually overthinking the problem. The back or the front of the train can be used as a marker. the back enters at the begining of 40 seconds and exits the tunnel after 40 seconds passes. The front enters the tunnel at the begining of 5 seconds and exits 5 seconds before the whole train exits or at 35 seconds or 5 seconds before 40.
Anyway the tunnel is 40 seconds long.
40 sec * 88 feet/sec = 3520 feet long