document.write( "Question 536239: A train is traveling at 22 m/s when it passes your town. Where will the train be 30 minutes later? \r
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\n" ); document.write( "\n" ); document.write( "Imagine that the train was 300 meters south of the your town when it was first observed. (We will assume that south is a negative direction). What is the trains POSITION at 50 seconds? How FAR did it travel in total though?
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Algebra.Com's Answer #352214 by fcabanski(1391)\"\" \"About 
You can put this solution on YOUR website!
Distance formula: D = rt where r is the rate (speed) and t is time.


\n" ); document.write( "In this case for part A:

\n" ); document.write( "D = what must be found.

\n" ); document.write( "r = 22 m/s

\n" ); document.write( "t = 30 minutes * 60 seconds/minute = 1800 seconds. (Convert minutes to seconds because the speed is given in meters per second.)

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\n" ); document.write( "\n" ); document.write( "D = 22*1800 = 39600 meters = 39.6 kilometers from town.


\n" ); document.write( "Part B: Use D=rt to find how far the train traveled in 50 seconds.

\n" ); document.write( "D = what must be found.

\n" ); document.write( "r = 22 m/s

\n" ); document.write( "t = 50 seconds.


\n" ); document.write( "D = 22 * 50 = 1100 meters.


\n" ); document.write( "The train began 300 meters south of town, which the problem specifies as -300. It moved 1100 meters north, which is +1100, so its position is now:


\n" ); document.write( "1100-300= 800. For the problem, +800 means 800 meters north of town.\r
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