document.write( "Question 170799: At 7:00 a.m., Joe starts jogging at 6 mph. At 7:10 a.m., Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 a.m.? \n" ); document.write( "
Algebra.Com's Answer #126119 by ankor@dixie-net.com(22740)\"\" \"About 
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At 7:00 a.m., Joe starts jogging at 6 mph. At 7:10 a.m., Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 a.m.?
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\n" ); document.write( "From the information given we can say:
\n" ); document.write( "Joe's travel time = \"1%2F2\" hr
\n" ); document.write( "Ken's travel time = \"1%2F3\" hr
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\n" ); document.write( "Let s = Ken's jogging speed
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\n" ); document.write( "When Ken overtakes Joe, they will have traveled the same distance
\n" ); document.write( "Write a distance equation from this fact; Dist = Time * speed
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\n" ); document.write( "Kens dist = Joe's dist
\n" ); document.write( "\"1%2F3\"s = \"1%2F2\"*6
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\n" ); document.write( "Multiply both sides by 6 to get rid of the denominators
\n" ); document.write( "2s = 3*6
\n" ); document.write( "2s = 18
\n" ); document.write( "s = 9 mph is Ken's speed
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\n" ); document.write( "Check solution by finding the dist:
\n" ); document.write( "\"1%2F3\"* 9 = 3 mi
\n" ); document.write( "\"1%2F2\"* 6 = 3 mi
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