
#1
Feb1407, 11:30 PM

P: 2,268

1. The problem statement, all variables and given/known data
A ferry captain wishes to travel directly across a river. A current of 4km/h is flowing and the ferry can travel at 8km/h. a) In what direction should the captain direct the ferry? b) What is the resultant velocity of the ferry as seen by someone standing on the riverbank? 2. Relevant equations 3. The attempt at a solution a) The ferry should be directed to travel 26.6 degrees up stream so that it will travel in a straight line.  same as the answers. b) I don’t get. It’s not sqrt(8^2+4^2) because it travels in a straight line now. Would it simply be 8km/h? The answers suggested 6.9m/s. 



#2
Feb1507, 12:00 AM

P: 1,351

cos(30)*8=6.9km/hr 



#3
Feb1607, 03:12 AM

P: 2,268





#4
Feb1607, 04:11 AM

P: 1,351

Ferry crossing
To overcome the current, is the direction of the ferry, having X ,I
a component of 4Km/hr upstream to offset the drift. Since........XI the total velocity is 8Km/hr sin (theta)=4/8. To observer ............XI ferry is bearing directly at themhe doesn't see the angle.............X,I the ferry must make with respect to the shoreline since...................X the net velocity upstream is a wash, ie=0. The component dircted toward the river is then cos(30)*8. Your approach using pythagorans theorum ok, but should be: 8=sqrt(4^2+Vapperent^2), squaring both sides V^2=6416 



#5
Feb1607, 03:11 PM

P: 2,268

I see, the key idea was that the ship is travelling at an angle of 30 degrees upstream. There is a horizontal component of 4km/h directly upstream and a vertical speed directly crossing the shore. This speed can be calculated by the pythagoras theorem sqrt(6416)=6.93km/h. Note that the book got a) wrong. The angle upstream should be different to the angle the ship would be tilted had it travelled directly straight. Because in one case, 8km/h was the hypotenus, in the other it was not.




#6
Feb1607, 08:47 PM

P: 1,351

Exactly, complain re the text, there is no excuse for these types of errors.



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