document.write( "Question 798926: Two carloads of friends are driving cross-country from Connecticut to San Francisco. One car will be traveling an average speed of 55 mph and the other car will be traveling an average speed of 65 mph. They want to be in the same spot at the end of the day after the second car does 6 hours of driving. How much of a head-start (distance-wise) should the slower car have before the faster car starts so they are in the same place at the end of the day? (after finding that-can you tell me how long time-wise the head start would have to be? \n" ); document.write( "

Algebra.Com's Answer #482411 by Stitch(470) $\"\"$ $\"About$

You can put this solution on YOUR website!
First lets see how far the second car will go in 6 hours while driving 65mph.
\n" ); document.write( " $\"%2865miles%2Fcross%28hour%29%29%2A+6cross%28hours%29+=+390miles\"$
\n" ); document.write( "----------------
\n" ); document.write( "Now lets see how far the first car will go in 6 hours while driving 55mph.
\n" ); document.write( " $\"%2855miles%2Fcross%28hour%29%29%2A+6cross%28hours%29+=+330miles\"$
\n" ); document.write( "----------------
\n" ); document.write( "Subtract the mileage of the two cars.
\n" ); document.write( " $\"390miles+-+330miles+=+60miles\"$
\n" ); document.write( "The slower car would need a 60 mile head start.
\n" ); document.write( "----------------
\n" ); document.write( "To calculated that time-wise, divide 60miles by 55miles/hour.
\n" ); document.write( " $\"%2860cross%28miles%29%2F%2855cross%28miles%29%2Fhour%29%29+=+1.091hours\"$
\n" ); document.write( "You can also turn the decimal places of the hour to minutes by multplying them by 60.
\n" ); document.write( "0.091 * 60 = 5.46
\n" ); document.write( "So the slower car would need about a one hour and 6 minute head start. \n" ); document.write( "

\n" );