Question

Two boats start together and race across a 60-km-wide lake and back. Boat A goes across at 60 km/h and returns at 60 km/h. Boat B goes across at 30 km/h, and its crew, realizing how far behind it is getting, returns at 90 km/h. Turnaround times are negligible, and the boat that completes the round trip first wins. (a) Which boat wins and by how much? (Or is it a tie?) (b) What is the average velocity of the winning boat?

Answer 1

Two boats start together and race across a 60-km-wide lake and back. Boat A goes across at 60 km/h and returns at 60 km/h. Boat B goes across at 30 km/h, and its crew, realizing how far behind it is getting, returns at 90 km/h. Turnaround times are negligible, and the boat that completes the round trip first wins.

(a) Which boat wins and by how much? (Or is it a tie?)

(b) What is the average velocity of the winning boat?

a) For boat A the round trip takes time:

$$t _ {1} = \frac {2 l}{v _ {1}} = 2 * \frac {6 0}{6 0} = 2 h$$

where $l = 60 \, \text{km}$ wide of the lake, $v_{1}$ - speed of boat A

For boat B it takes time:

$$t _ {2} = \frac {l}{v _ {2 1}} + \frac {l}{v _ {2 2}} = \left(\frac {6 0}{3 0} + \frac {6 0}{9 0}\right) h = 2, 6 6 h$$

where $v_{21} = 30\frac{km}{h}$ and $v_{22} = 90\frac{km}{h}$ speeds of boat B

So, the boat A completes the round trip first. The difference of times equals:

$$\Delta t = t _ {2} - t _ {1} = 2, 6 6 h - 2 h = 0. 6 6 h = 4 0 \text{ min}$$

Answer: boat A, $\Delta t = 40$ min

b) Average speed for boat A equals:

$$v _ {a} = \frac {2 l}{t _ {1}} = \frac {1 2 0}{2} \frac {k m}{h} = 6 0 \frac {k m}{h}$$

(and because boat A has constant speed during all trip)

Answer: $60\frac{km}{h}$