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rdn98

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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.

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?

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?

--------

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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