Cycling word problem

Question

as much as I try I can not find any equation that would give me the correct answer to this problem.

 

"a cyclist leaves his training base for a morning workout at the rate of 18 mph. One hour later, his support staff leaves the base in a car going 45 mph in the same direction. How long will it take the support staff to catch up to the cyclist?"

Answer 1

Rate*Time = Distance

 

Time = Distance/Rate

 

The cyclist has an 18 mile headstart.  If his staff were traveling at 18 mph, the 18 mile gap would never get smaller. and they'd never catch up with the cyclist.   But they are traveling at 45 mph, 45-18=27 mph faster than the cyclist, so they are closing the 18 mile gap at 27 mph:

 

Time = (18 miles) / (27 mph) = ____________?

 

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Another way to do it:

 

The distance the cyclist has traveled in time t is given by the linear equation y = 18t

 

The equation for the staff is y = 45t - 18  (at t=0, they are 18 miles behind the cyclist, or -18).  The staff catches up with the cyclist when they have traveled the same distance:

 

45t - 18 = 18t

27t = 18

 

t = ____?