document.write( "Question 596745: John started walking East at 10:00 AM at a rate of 2 km/hr. James stated walking South at 10:30 AM at a rate of 4km/hr. What time is it when they are 5 km apart?\r
\n" ); document.write( "\n" ); document.write( " D (x) = Rate x time
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\n" ); document.write( "\n" ); document.write( "Tried formula below, but stuck on time part????\r
\n" ); document.write( "\n" ); document.write( "a2 + b2 = c2 so 4x2 + 2x2 = 25 \r
\n" ); document.write( "\n" ); document.write( "x= 1.25\r
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Algebra.Com's Answer #377856 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "At 10:30 when James starts, John will be 30 minutes into his trip. 30 minutes at 2 km/hr is 1 km. One hour later John will be 3 km from the start and James will be 4 km from the start, making a 3-4-5 right triangle at time 11:30.\r
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\n" ); document.write( "\n" ); document.write( "That was the easy way. Now the hard way:\r
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\n" ); document.write( "\n" ); document.write( "In the first 30 minutes John goes 1 km. Every hour after that, he goes 2 km. So counting hours since JAMES starts, John goes km. James is going twice as fast as John, so every hour after James starts he goes km.\r
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\n" ); document.write( "\n" ); document.write( "In order for them to be 5 km apart:\r
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\n" ); document.write( "\n" ); document.write( "Simplifying:\r
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\n" ); document.write( "\n" ); document.write( "Solve for the . Toss out the negative value (you don't care what happened before either of them started the trip, do you?)\r
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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