vegetars8t

2022-01-05

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. What is the average velocity of the winning boat?

Step 1
Here width of the lake
$w=60km$
Up speed of the boat-1
Down speed of the boat-1
Up speed of the boat-2
Down speed of the boat-2
The time taken by the boat-1 for the trip is
${t}_{1}=\frac{w}{{v}_{1U}}+\frac{w}{{v}_{1D}}$
1)
$=2h$
And the time taken by the boat-2 for the trip is
${t}_{2}=\frac{w}{{v}_{2U}}+\frac{w}{{v}_{2D}}$
2)
$=2.666h$
$=2.7h$
Because from equations (1) and (2), we can see that
${t}_{2}>{t}_{1}$
Therefore boat-1 wins the trip by 0.7 h
Because the winning boat-1 returns at the same point from where it stars. Therefore its displacement is zero. So the average velocity of the winning boat-1 is
$\stackrel{\to }{\stackrel{―}{v}}=\frac{\text{displacement}}{\text{time taken}}$
$=\frac{0}{2h}$

nghodlokl

Step 1
Average speed for boat A equals:
${v}_{a}=\frac{2l}{{t}_{1}}=\frac{120}{2}\frac{km}{h}=60\frac{km}{h}$
(and because boat A has constant speed during all trip)

karton

Step 1
$\text{Average Velocity}=\frac{\text{Total displacement}}{\text{time taken}}$
$\text{Time taken}=2hr$
$\text{Total displacement}=0km,$
Since boat A starts at starting point of lake and returns back after completing race at the same starting point, So total displacement is zero.
So, $\text{Average velocity}=\frac{0km}{2hr}=\frac{0km}{hr}$

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