A cyclist leaves his training base for a morning workout, riding at the rate of 18 mph. One hour later, his support staff leaves the base in a car going 52 mph in the same direction. How long will it take the support staff to catch up with the cyclist once the car leaves the base?

Answer 1

At the one hour point the cyclist is 18 miles from from the training base and his support staff.
From this point on, his support staff will be closing in on the cyclist at the closing speed of 52 - 18 = 34mph.
Thje initial gap of 18 miles will therefore be closed in 18/34 = .514285 hours or 30min - 51.42sec.

Answer 2

Since car #1 has a 1 hr head-start,
d1 = 18 mi/hr * 1hr + 18t,
d1 = 18 + 18t,
d2 = 52t,
When car #2 catches up:
d1 = d2,
18 + 18t = 52t,
18 = 52t - 18t = 34t,
t = 18 / 34 = 0.5294 hrs = 31.8 min