Question

Additional Practice on Speed

1. A group of campers and one group leader left a campsite in a canoe.

They traveled at an average rate of 10 km/h. Two hours later, the other

group leader left the campsite in a motorboat. He traveled at an average

rate of 22 km/h. How long had each boat traveled when the motorboat

caught up with the canoe?

a.

canoe: 2.2 h; motorboat: 1 h

b. canoe: 2.2 h; motorboat: 10 h

c. O canoe: 3 2 h; motorboat: 12 h

3

3

d. O canoe: 4.4 h; motorboat: 2 h

Answer 1

canoe = 3 h 40 min. motorboat = 1 h 40 min

Step-by-step explanation:

• As both boats travel at a constant speed, we can apply the definition of average velocity to both, so we can get:

`v_(b) = (\Delta x_(b) )/(\Delta t_(b) ) (1)`

`v_(mb) = (\Delta x_(mb) )/(\Delta t_(mb) ) (2)`

• Now, when the motorboat caught up with the canoe, distance travelled by both boats were the same, so, solving (1) and (2) for Δx, we get:

`v_(b) *\Delta t_(b) = v_(mb) *\Delta t_(mb) (3)`

• if we take the initial time for the canoe to be t₀ =0, the initial time for the motorboat will be tmb₀ = 2 Hs.
• So, replacing in (3) we have:

`v_(b) *t_(b) = v_(mb) *(t_(b) - 2 hs.) = 10 km/h* t_(b) = 22 km/h* (t_(b) - 2 hs.)`

• Solving for tb:

`t_(b) = (-44 )/(-12) hs = 3h 40 min.`

tmb = tb - 2 hs. = 1h 40 min.