A river 570 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.(adsbygoogle = window.adsbygoogle || []).push({});

a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?

b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. If she heads 35° upstream, how far downstream is she swept before reaching the opposite bank?

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

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I'm stuck on part b only. :-(

Here are some equations I figured out from my book.

xf= t * vx = ( d / vy )* vx

t=d/vy

vy = vA*cos (theta)

vx = vA*sin(theta) + vC

where vA is the velocity of the swimmer

Vc is the velocity of the river

vy is the y component of velocity

vx is the x component of velocity

xf is final position

t=time

So for part a) I did position=(d/vy)*(vx) and the angle is zero, so

(570ft/3ft)(6ft)= 1140 ft, which is right.

Part c)

I did:

vy=vA*cos theta

vy= 3*cos 35 =2.457 m/s

then t= d/vy => (570ft)/(2.457 m/s)= 231 seconds.

Now for part b:

I tried this:

vx=vA*sin theta+vC

vx=3*sin 35 + 6 =>7.720

vy=3*cos 35 =>2.457

position=(d/vy)*(vx)=> (570/2.457)(7.720)=1790.96 ft

Thing is, computer won't take my answer, and I believe my answer is correct. Hopefully there isn't a bug in the computer program. Is there another way of doing this problem, maybe easier?

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# Homework Help: Swimmer in river relative motion crap

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