Homework Help: Swimmer in River

1. Apr 8, 2005

Naeem

Q. A river 590 ft wide flows with a speed of 6 ft/s with respect to the earth. A woman swims with a speed of 3 ft/s with respect to the water.
--------------------------------------------------------------------------------
a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
d1 = ft *
1179.96 OK

--------------------------------------------------------------------------------
b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. If she heads 29° upstream, how far downstream is she swept before reaching the opposite bank?
d2 = ft

Any ideas on part b

--------------------------------------------------------------------------------
c) For the conditions of part (b), how long does it take for her to reach the opposite bank?
Any ideas on part c

2. Apr 8, 2005

Quantum Cat

Aha I remember this from vectors in grade 10.

Remember relative velocity law:

V woman,earth = V woman,water + V water,earth

Draw your diagrams, look at the components, add them. find scalar multiple k so that the vertical component is 590 (the width of river) and the horizontal component is how far she is swept downstream. k is the time it takes to get the the other side.

Part b: exactly the same as part a, except change your vectors so she heads 29 degrees instead of being perpendicular to the current of the river.

Part c: same as part a; find k a scalar multiple such that the vertical component is the width of the river. k is the time it takes.

By the way it's so funny seeing things in feet and feet per second. I kind of gained an impression about how annoying it must be for poor americans learning science and having been brought up to know the old-fashioned units and not metric!!

3. Apr 9, 2005

Naeem

Here is what I did for part b,

d = v0 cos theta * t

width of the river, d = 590 m

velocity of the river = 6ft /s

velocity of the woman = 3ft/s

so, t = d / vwoman * cos (29)

t= 590 /3 * cos(29)

After finding t , we can multiply this t with 6 ,which is the velocity of the river.