Finding boat speed when Mike and Tina race to a sauna

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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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Discussion

Is 1.104 supposed to be the distance from A to C? It doesn't look like 6.9 * 1/6.
 
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.