Question 262305
 A man jogs at 6 mph while a cyclist moves at 18 mph. After how long will the jogger be overtaken if the cyclist leaves one hour after the jogger begins the course? 
This is the table that I've set up so far, I cannot seem to get beyond this point because I do not know a distance for the two travelers. 
Speed: 6 18
Distance: 
Time: x x+60
<pre><font size = 4 color = "indigo"><b>
You don't need to change 1 hour to minutes. You wouldn't
want to do that, because if you did that you'd have to change'
the speeds to miles per minute instead of miles per hour and you
wouldn't want to do that!

First, let's do it in our head, then we'll do it by algebra.

Here's how to do it in your head: 

When the cyclist starts, the jogger is already 6 miles down the road.
So the jogger has a 6-mile head start on the cyclist. The cyclist goes
18mph and the jogger goes 6mph, so the cyclist approaches the jogger
at a rate of 18mph-6mph or 12mph.  So, approaching the jogger at 12mph,
he can make up the jogger's 6 mile head-start in half an hour. 

Now here's the algebra way:

Make this chart:

             DISTANCE      RATE      TIME
jogger        
cyclist        

You want to know the cyclist's time, so put t for that


             DISTANCE      RATE      TIME
jogger        
cyclist                                t

The jogger's time was 1 hour more so add 1 to t and put
t+1 for the jogger's time:

             DISTANCE      RATE      TIME
jogger                                t+1
cyclist                                t


Now fill in their rates which are given:

             DISTANCE      RATE      TIME
jogger                       6        t+1
cyclist                     18         t


Now use DISTANCE = RATE x TIME to fill in the
distances.

             DISTANCE      RATE      TIME
jogger         6(t+1)        6        t+1
cyclist         18t         18         t


Now when the cyclist catches up to the jogger,
they will have both traveled the same distance,
so we set the expressions for their distances equal
to each other:

      6(t+1) = 18t
      6t + 6 = 18t

Subtract 6t from both sides

           6 = 12t

Divide both sides by 12

          {{{6/12=12t/12}}}

Cancel the 12's on the right, and
reduce the fraction on the left:

          {{{1/2=t}}}

So the answer is half an hour, just what we got
when we did it in our heads.

Edwin</pre>