Question

A group of campers and one group leader left a campsite in a canoe. They traveled at an average rate of10km/h. Two hours later, the other group leader left the campsite ina motorboat. He traveled at an average rate of 22km/h.
a. How long after the canoe left the campsite did the motorboat catch up with it?
b. How long did the motorboat travel?

Answer 1

a 2 hr head start for the canoe is a 20 km lead

the boat closes the gap at 12 kph (22 - 10)

a. 2 hr + (20 km / 12 kph)

b. 20 km / 12 kph

Answer 2

Speed:

v = s / t

s = distance

t = time

t = s / v

t1 = travel time of group with canoe

t2 = travel time of group with motorboat

t1 = s / v1

t1 = s / 10

t2 = 2 hours + s / 22

t2 = 2 + s / 22

The motorboat catch up canoe when:

t1 = t2

s / 10 = 2 + s / 22

Subtract s / 22 to both sides

s / 10 - s / 22 = 2

11 ∙ s / 11 ∙ 10 - 5 ∙ s / 5 ∙ 22 = 2

11 s / 110 - 5 s / 110 = 2

6 s / 110 = 2

Multiply both sides by 110

6 s = 220

Divide both sides by 6

s = 220 / 6 = 2 ∙ 100 / 2 ∙ 3 = 110 / 3 km = 36.666 km

t = t1 = s / v1

t = ( 110 / 3 ) / 10 = 110 / 30 = 10 ∙ 11 / 10 ∙ 3 = 11 / 3 = 9 / 3 + 2 / 3 = 3 + 2 / 3

3 h + 2 h / 3 = 3 h + 2 ∙ 60 min / 3 = 3 h + 2 ∙ 20 min = 3 h 40 min

OR

t = t2 = 2 + s / 22 = 2 + ( 110 / 3 ) / 22 = 2 + 110 / 66 =

2 + 22 ∙ 5 / 22 ∙ 3 = 2 + 5 / 3 = 2 + 3 / 3 + 2 / 3 = 2 + 1 + 2 / 3 = 3 + 2 /3

= 3 + 2 ∙ 60 min / 3 = 3 h + 2 ∙ 20 min = 3 h 40 min

a.

3 h 40 min

b.

110 / 3 km = 36.666 km