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A swimmer wants to cross a river, from point A to point B, as shown in the figure. The distance d1 (from A to C) is 200 m, the distance d2 (from C to B) is 150 m, and the speed vr of the current in the river is 5 km/hour. Suppose that the swimmer's velocity relative to the water makes an angle of θ=45degrees with the line from A to C, as indicated in the figure.

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A) To swim directly from A to B, what speed us, relative to the water, should the swimmer have?

Express the swimmer's speed numerically, to three significant figures, in kilometers per hour.

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Added Wed, 06 Apr '16
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We know that the time taken to move 200m in the y direction (d1) has to be the same time taken to go 150m in the x direction (d2). This means the swimmer is essentially moving along the hypotenuse of the triangle. We can set up a system of equations to solve the problem:

x-x0 = vxt + 1/2axt2

There�s no acceleration so that part cancels out. The swimmer is swimming to the left but the current is pulling him to the right � we�ll treat the swimmer�s velocity as positive and the current�s velocity as negative. We know that the swimmer needs to move 150m to the right � again, from the swimmer�s perspective, this is negative. Also, because ? is with respect to the y direction, we have to use sin to find the x velocity and cos to find the y velocity (there are other ways of solving the problem but this is the most direct one). We end up with:

-150 = (sin(45)v � 5000)t

In other words, the swimmer�s x velocity is a combination of his own swimming plus the negative effect of the current. Now let�s set up an equation for the y distance, but put it in terms of t so that we can plug it into the above equation:

y-y0 � vyt + 1/2ayt2

Again, no acceleration (the swimmer is moving up, but not against gravity), so that part cancels. We�re left with:

200 = cos(45)vt

Solving for t, we get:

t = 200/(cos(45)v)

t = 200/(0.7071v)

Now plug this into the first equation:

-150 = (sin(45)v � 5000)t

-150 = (0.7071v � 5000)(200/(0.7071v))

-150(0.7071v/200) = (0.7071v � 5000)

-0.530v = 0.7071v � 5000

1.2374v = 5000

v = 4040.7306m/h

Convert to km/h like the problem asks for:

v = 4.04km/h

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Added Wed, 06 Apr '16
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