Question

A group of campers and one group leader left a campsite in a canoe traveling at a steady 8 km/h. One hour later, the other group leader left left the campsite in a motorboat with the supplies. The motorboat followed the canoe at a steady 20 km/h. How long after the canoe left the campsite did the motorboat overtake it. Show all the work..

A. 2 1/2 h
B. 1 2/3 h
C. 3/4 h
D. 2/5 h

Answer 1

B. 1 2/3 h

Explanation:

Let t represent the time since the canoe left the campsite. The distance it covers is ...

distance = rate · time

distance = 8t

The time the motorboat travels is t-1 (1 hour less time than the canoe). The distance it covers is ...

distance = 20(t -1)

The two distances are equal at the time of interest, so we have ...

20(t -1) = 8t

20t -20 = 8t . . . . . eliminate parentheses

12t = 20 . . . . . . . . . add 20-8t to both sides

t = 20/12 = 5/3 = 1 2/3 . . . . hours after the canoe left

The motorboat overtakes the canoe 1 2/3 hours after the canoe left.

_____

Additional comment

The domain of the motorboat's distance function is t ≥ 1. It doesn't start moving until after the canoe has been gone for an hour.

The time after the motorboat starts is 2/3 hour. That is the time it takes to cover the initial difference in distance at a speed that is the difference in speeds: 8 km/(12 km/h) = 2/3 h. This problem asks for the time since the canoe started, so we need to add an hour to that.