# Whered i go wrong?

1. Sep 15, 2004

### egg man

a river flows at a speed Vr=5.84 km/hr with respect to shore. a boat needs to go perpendicular to the shoreline to reach a pier across the river. the boat heads upstream at an angle of 30 degrees. if the time taken for the boat to cross is 15.3 min, what is the width of the river?

okay, i know the actual velocity of the boat over the velocity of the river(5.84) equals 1/sin30. so the velocity of the boat equals 11.68 km/hr. i divide by 60 and i get .195km/min. i multiply by 15.3 and get 2.98km. but thats not right. Pleeaaaase help

2. Sep 15, 2004

### HallsofIvy

Staff Emeritus
What I would do is break the vector velocities into "components". If the boat is aimed at 30 degrees (to the perpendicular) with speed Vr, then the x and y components of the velocityh vector are Vr cos(30) and Vr sin(30) respectively.
At that speed, the distance he goes (across the river- ignore the up or downstream motion) is Vr cos(30)*t= (5.84)(cos(30))(15.3).

Notice that the speed of the river doesn't come into this! That would, of course, affect what angle we have to aim upstream but we are, apparently, told that this was 30 degrees.