Question

A man starts out on a trip driving his car at a constant rate of 48 mph. A man riding a motorcycle starts out on the same route 1 1 2 12 hours later, traveling at a constant rate of 64 mph. How long will the car have been traveling when the motorcycle catches up?

Answer 1

6 hours

Explanation:

The car has a head start of (48 mi/h)(1.5 h) = 72 mi. The motorcycle is catching up at the rate of (64 -48) = 16 mi/h. At that rate, the motorcycle will catch up to the car in (72 mi)/(16 mi/h) = 4.5 h.

When the motorcycle catches the car, it will have been traveling 4.5 hours. The car traveled 1.5 hours longer. The car will have been traveling for 6 hours.

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Check

They will meet at a distance of 48·6 = 288 = 64·4.5 miles from the start.

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Alternate Solution

If you like, you can write an equation for the distance covered from the time the car starts:

car's distance = 48t

motorcycle's distance = 64(t -1.5)

These distances are equal when ...

48t = 64t - 96

96 = 16t . . . . . . . add 96-48t

96/16 = 6 = t . . . . the motorcycle catches up 6 hours after the car starts.

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Note that in this solution, we are solving for the time from the car starting. In the above "word" solution, we solved for the time from the motorcycle starting, then added the time before the motorcycle started.