The Lake House: Mike and Tina's Race to the Sauna

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Homework Help Overview

The problem involves calculating the average speed of a boat traveling from an island's beach house to a lakeside sauna, with a comparison to a person's walking speed along the beach. The context includes distances and times related to both Mike and Tina's journeys.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the distances involved, questioning the calculations and assumptions about the distances from points A to C and B to C. There are varying interpretations of the distances based on different rounding methods and the application of the Pythagorean theorem.

Discussion Status

The discussion is ongoing with multiple interpretations of the distances and speeds being explored. Some participants provide calculations while others question the accuracy of those calculations and the terminology used, such as the term "average speed."

Contextual Notes

There is confusion regarding the distances calculated and the implications of rounding in the context of the problem. Participants are also considering the definitions of average speed versus instantaneous speed.

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Homework Statement


The lake house is located in A, which is home of Mike. Against the shore, perpendicular to the 675 m distance is the island's beach house B, which is home of Tina. Mike walks along the beach at a speed of 6.9 km/h toward the sauna and the C arrives there exactly 10 minutes. At the same time Tina also leaves the island by boat lakeside sauna and the C arrives there exactly 4 minutes earlier than Mike.

Homework Equations


What was the average speed of the boat?


The Attempt at a Solution


x2=0.6752+1.1042
x=1.294Km (distance from B to C)

Tina arrives 4minutes earlier than Mika means she arrives in 10-4=6min=0.1h
Average speed = 1.294/0.1=12.94km/h=3.594m/s
 

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Is 1.104 supposed to be the distance from A to C? It doesn't look like 6.9 * 1/6.
 
yes it is.
distance from A to C = 6.9*0.16=1.104Km
 
chawki said:
yes it is.
distance from A to C = 6.9*0.16=1.104Km

Interesting about rounding. 10 min/ 60 min is 0.16666etc., which rounds to 0.17 (have to round up), producing a distance of 1.173 km.

Simply dividing by 6 (*10/60) gives 1.15 km exactly.

You seem to be doing the rest the right way.
 
I don't know where you got 1.15km or what it is!

by using the phytagorean law, i found the distance from B to C is 1.353km.
Then the speed would be 1.353/0.1 = 13.53km/h

but I'm still wondering why they said average speed..obviously it's just a speed.
 
chawki said:
I don't know where you got 1.15km or what it is!

by using the phytagorean law, i found the distance from B to C is 1.353km.
Then the speed would be 1.353/0.1 = 13.53km/h

but I'm still wondering why they said average speed..obviously it's just a speed.

1.15km is the distance A to C. You are using 1.104 based on improper rounding. I didn't bother to calculate how much this affects your result.
 
no i used 1.173km