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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 :

cjrodriguez89

Answer:

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

Explanation:

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

        [tex]v_{b} = \frac{\Delta x_{b} }{\Delta t_{b} } (1)[/tex]

       [tex]v_{mb} = \frac{\Delta x_{mb} }{\Delta t_{mb} } (2)[/tex]

  • 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:

        [tex]v_{b} *\Delta t_{b} = v_{mb} *\Delta t_{mb} (3)[/tex]

  • 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:

        [tex]v_{b} *t_{b} = v_{mb} *(t_{b} - 2 hs.) = 10 km/h* t_{b} = 22 km/h* (t_{b} - 2 hs.)[/tex]

  • Solving for tb:

        [tex]t_{b} = \frac{-44 }{-12} hs = 3h 40 min.[/tex]

        tmb = tb - 2 hs. = 1h 40 min.

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