Question 433975
<pre>
A long distance runner started on a course running at an average speed of 6 mph. one-half hour later, a second runner began the
same course at an average speed of 7mph.  How long after the second runner started did the second runner overtake the first runner?

Let time taken by FASTER runner to get to meet up point, be T
With FASTER runner's speed being 7 mph, DISTANCE FASTER runner covers, to get to meet-up point = 7T 

Since SLOWER runner started one-half ({{{1/2}}})hour before FASTER runner, then time slower runner takes to get to meet-up point = T + {{{1/2}}}
With SLOWER runner's speed being 6 mph, DISTANCE SLOWER runner covers, to get to meet-up point = 6(T + {{{1/2}}}) 

When both got to meet-up point, they'd covered the same distance
As such, we get the following DISTANCE equation: 7T = 6(T + {{{1/2}}})
                                                 7T = 6T + 3
                                            7T - 6T = 3
<font color = red><font size = 4><b>Time FASTER runner takes to get to meet-up point/catch up with SLOWER runner</font></font></b>, or T = <font color = red><font size = 4><b>3 hours</font></font></b></pre>