```Question 88072
I have to solve this proble and show equation but I do not
know where to start could you help me please?
Martina leaves home at 9 A.M., bicycling at a rate of 24 mi/h.
Two hours later, John leaves, driving at the rate of 48 mi/h.
At what time will John catch up with Martina?
<pre><font face = "lucida console" size = 5 color = "brown"><b>
You can do this in your head, but I'll do it by
algebra as well.

at 11AM, when John leaves, Martina is already
48 miles down the road, since she has pedaled
at the (break-neck) speed of 24 mi/hr on a
bicycle for two hours!  So John, driving at
48 mi/hr is approaching her at 24 mi/hr, so he
has to make up her 48 mile lead at 24 mi/hr
approach rate, so it'll obviously take him 2
hours, and 2 hours past 11AM is 1PM.

How to do it by algebra.  Make this chart:

DISTANCE     RATE     TIME
Martina
John

Let the answer be t. So write t for John's
catch-up time

DISTANCE     RATE     TIME
Martina
John                                t

Since Martina has a 2-hour head start, her
time is t+2. So fill that in:

DISTANCE     RATE     TIME
Martina                            t+2
John                                t

Now fill in their rates which are given as
24 mi/hr and 48 mi/hr

DISTANCE     RATE     TIME
Martina                   24       t+2
John                      48        t

Now use DISTANCE = RATE×TIME to fill in the
two DISTANCEs:

DISTANCE     RATE     TIME
Martina      24(t+2)      24       t+2
John           48t        48        t

Now since they both traveled the same distance,
we set the DISTANCEs equal:

24(t+2) = 48t

Solve that and get t = 2 hours, which means John,
starting two hours later than 9AM, or 11AM, will
catch up to her in 2 hours past 11AM, or 1PM.

Edwin</pre>

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