SOLUTION: Question: A group of campers and one group leader left a campsite in a canoe traveling at a steady 8km/h. One hour later, the other gruop leader left the campsite in a motorboat

Question 417707: Question: A group of campers and one group leader left a campsite in a canoe traveling at a steady 8km/h. One hour later, the other gruop leader 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 motorbat overtake it?

This is what I did so far:
Canoe: 8 km/h.
1st Hour: 8
2nd Hour: 16
motorboat: 20 km/h. (with a delay of 1 hour after canoe)
1st Hour: 0
2nd Hour: 20
Hence, it can be concluded right away that it will take no less than 1 hour nor over 2 hour to overtake the canoe.
Now this is where I stuck.
so, I used the ratio of (16/20) to find the difference.
Hence, the answer comes out as 1.8
The problem is that the book told me the answer suppose to be 1(2/3) h. and not 1(4/5) h.
Can you please show me the right way to do it. If possible, I want to know the proper equation to use with these type of question.
Thank you very much.
From,
Wolf Schneider

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Use distance equals rate times time.

Let

be the distance traveled by the canoe at the time the motorboat catches up. Let

be the amount of time the canoe was travelling. And of course the rate of the canoe is given as 8 km/hr. So we can describe the trip taken by the canoe as:

Since both the canoe and the motorboat start from the same place and they are at the same place when the motorboat catches up, they must have traveled the same distance. Hence we can let

represent the distance traveled by the motorboat. Since the motorboat left 1 hour later, the elapsed time of the motorboat trip is