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?

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Nadine Salcido · Accepted Answer

Step 1  Here width of the lake  w=60km  Up speed of the boat-1 v1U=60km h−1  Down speed of the boat-1 v1D=60km h−1  Up speed of the boat-2 v2U=30km h−1  Down speed of the boat-2 v2D=90km h−1  The time taken by the boat-1 for the trip is  t1=wv1U+wv1D  1) =60km60km h−1+60km60km h−1  =2h  And the time taken by the boat-2 for the trip is  t2=wv2U+wv2D  2) =60km30km h1+60km90km h−1  =2.666h  =2.7h  Because from equations (1) and (2), we can see that  t2&amp;gt;t1  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  v―→=displacementtime taken  =02h  =0m s−1

nghodlokl · Accepted Answer

Step 1    Average speed for boat A equals:   va=2lt1=1202kmh=60kmh    (and because boat A has constant speed during all trip)

karton · Accepted Answer

Step 1 Average Velocity=Total displacementtime taken Time taken=2hr 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, Average velocity=0km2hr=0kmhr

Two boats start together and race across a 60-km-wide lake and back. B

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