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Exuro89

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## Homework Statement

A 400m wide river flows from west to east at 30 m/min. Your boat moves at 100m.min relative to the water no matter which direction you point. To cross the river, you start from a dock at point A on the south bank. There is a boat landing directly opposite at point B on the north bank, and also one at point C, 75m downstream from B.

(a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled?

(b)If you initially aim your boat directly toward point C and do not change the bearing relative to the shore, where on the north shore will you land?

(c) To reach point C:(i)at what bearing must you aim you boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) what is the speed of your boat as measured by an observer standing on the river bank?

[PLAIN]http://i.min.us/jcg2eG.png [Broken]

## Homework Equations

X = X_0+V_0*t

X^2 + Y^2 = R^2

cos(theta)=opp/hypo

sin(theta)=adj/hypo

## The Attempt at a Solution

I'm stuck at C but I'll show you what I did for parts A and B to make sure that I'm attempting the problem correctly.

(a)What I did for the first part was find the time it take to go across the river. I found this by using

Y = Y_0 + V_0y*t

400m = 0m + 100m/min * t

400m/100m/min = t

t = 4min

Because the river is the only x component affecting the boat, I can take that and multiply that by 4 minutes to get the distance on the north shore.

X = X_0 + V_0x*t

X = 0 + 30m/min*4min

X = 120m

The second part of A asks the distance traveled. Because I have both X and Y distance I can use the pythagorean theorem to find the distance traveled.

X^2 + Y^2 = R^2

120^2 + 400^2 = R^2

R = 417.6m

(b) So for this question the boat is initially going towards point C, so I can find the degree right of the y axis because I have both Y length 400, and X length 75.

tan(theta) = 75/400

theta = tan^-1(75/400)

theta = 10.6 degrees

I can take theta and find both components of V_0

V_0y = 100cos(theta)

V_0y = 98.3m/min

V_0x = 100sin(theta)

V_0x = 18.4m/min

After this I can find how long it take to get across

Y = Y_0 + V_0y*t

400m = 0 + 98.3m/min*t

t = 4.07min

With the time I can multiply it by the x component(along with the added river's x component) to get the x distance

X = X_0 + V_0x*t

X = 0 + (18.4m/min+30m/min)*4.07min

X = 197m

So by pointing towards point C initially you would go 197m east of point C.

(c)Okay I'm having trouble with this. I'm not quite sure how I would find the bearing. I tried this, as I don't know either theta or time, i figured I could figure out 2 equations for time and then equal them to each other, but Im stuck.

Y = Y_0 + V_0y*t X = X_0 + V_0x*t

400m = 0 + 100m/mincos(theta)*t 75m = 0 + (100m/minsin(theta)+30m/min)*t

I solved for t, made them equal, but can't get theta out. Am I doing this wrong, what should I try, and are the earlier answers correct?

Thanks.

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