document.write( "Question 1056047: Janet set off on a bicycle trip at 630am at an average rate of 26 miles per hour. her husband . Bob, promised to bring her some lunch.
\n" ); document.write( "he set off by car 2 hours after janet left driving at an average speed of 40 miles per hour. At what time did bob catch up with Janet? \n" ); document.write( "

Algebra.Com's Answer #671210 by josgarithmetic(39630) $\"\"$ $\"About$

You can put this solution on YOUR website!
Time $\"t=0\"$ begins at 6:30AM, but time t will be some other, positive value and this quantity can be easier to use than a time point on a line.
\n" ); document.write( "Bob's catchup distance is d.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "NOW, $\"t%3C%3E0\"$ and you will solve for it.
\n" ); document.write( "

\r\n" );
document.write( "                    SPEED        TIME      DISTANCE\r\n" );
document.write( "JANET                26          t+2        d\r\n" );
document.write( "BOB                  40           t         d\r\n" );
document.write( "

\n" ); document.write( "

\n" );