A fisherman sets out upstream from Metaline Falls on the Pend Oreille River in northwestern Washington State. His small boat, powered by an outboard motor, travels at a constant speed v in still water. The water flows at a lower constant speed vw. He has traveled upstream for 1.65 km when his ice chest falls out of the boat. He notices that the chest is missing only after he has gone upstream for another 12.5 minutes. At that point, he turns around and heads back downstream, all the time traveling at the same speed relative to the water. He catches up with the floating ice chest just as it is about to go over the falls at his starting point. How fast is the river flowing? Solve this problem in two ways.

(a) First, use the Earth as a reference frame. With respect to the Earth, the boat travels upstream at speed v-vw and downstream at v+vw.

(b) A second much simpler and more elegant solution is obtained by using the water as the reference frame. This approach has important applications in many more complicated problems, such as calculating the motion of rockets and satellites and analyzing the scattering of subatomic particles from massive targets.

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(a) Using the Earth as a reference frame:

Let's denote the speed of the boat in still water as v, and the speed of the river's flow as vw.

When the fisherman is traveling upstream, his effective speed is reduced by the speed of the river flow. So, his speed relative to the Earth is v - vw.

When the fisherman is traveling downstream, his effective speed is increased by the speed of the river flow. So, his speed relative to the Earth is v + vw.

Given that the fisherman noticed his ice chest missing after traveling upstream for 1.65 km, and then it took him another 12.5 minutes to turn around, we can determine the time it took for him to travel the 1.65 km upstream.

Using the equation: distance = speed × time, we can write:

1.65 km = (v - vw) × t, where t is the time taken to travel the 1.65 km upstream.

Converting the distance to meters and the time to seconds, we have:

1650 m = (v - vw) × t, where t is now in seconds.

Next, we can determine the time it took for the fisherman to catch up with the ice chest when traveling downstream. Since he caught up with it just as it was about to go over the falls, the distance traveled by both the boat and the ice chest must be the same.

Using the equation: distance = speed × time, we can write:

1650 m = (v + vw) × t', where t' is the time taken to catch up with the ice chest.

Converting the distance to meters, we have:

1650 m = (v + vw) × t', where t' is in seconds.

Now, we have two equations:

1650 = (v - vw) × t (Equation 1)
1650 = (v + vw) × t' (Equation 2)

Dividing Equation 1 by Equation 2, we can eliminate the variable v, and solve for vw:

(v - vw) × t / ((v + vw) × t') = 1.

Simplifying this equation, we get:

t / t' = (v + vw) / (v - vw).

Given that t' is 12.5 minutes, which is equivalent to 750 seconds, we can write:

t / 750 = (v + vw) / (v - vw).

Now, we can rearrange this equation to solve for vw:

(v + vw) / (v - vw) = t / 750.

Cross-multiplying, we get:

(v + vw) × 750 = (v - vw) × t.

Expanding the equation, we have:

750v + 750vw = vt - vwt.

Rearranging terms, we get:

vt + vwt = 750v + 750vw.

Factoring out v from the left side, we have:

v(t + wt) = 750(v + vw).

Finally, simplifying the equation, we get:

t + wt = 750.

Substituting the given values for t and w, we have:

1.65 + (1.65 / vw) × 1.65 = 750.

Solving this equation will give us the value of vw, which represents the speed of the river flow.

(b) Using the water as the reference frame:

In this approach, we consider the water as a stationary frame of reference. Since the boat is traveling at the same speed relative to the water both upstream and downstream, we can equate the distances traveled to determine the speed of the river flow.

When the fisherman is traveling upstream, he travels for a certain time t' before realizing the chest is missing. During this time, the ice chest has been carried downstream by the river flow. The distance traveled by the chest during this time t' is vw × t’.

When the fisherman turns around and travels downstream, he catches up with the chest just as it is about to go over the falls at his starting point. The distance traveled by both the boat and the chest downstream is the same, which is the distance from the point where the chest fell to the starting point.

This distance is the same as the distance the fisherman traveled upstream before realizing the chest was missing, which is 1.65 km or 1650 m.

Therefore, we can write the equation:

Distance traveled by chest downstream = Distance traveled upstream by the fisherman

vw × t' = (v - vw) × t.

Dividing both sides by vw, we get:

t' = (v - vw) × t / vw.

Since t' is 12.5 minutes or 750 seconds, we can write:

750 = (v - vw) × t / vw.

Cross-multiplying, we have:

750vw = (v - vw) × t.

Expanding the equation, we get:

750vw = vt - vwt.

Rearranging terms, we have:

750vw + vwt = vt.

Factoring out v, we get:

v(750w + wt) = vt.

Finally, simplifying the equation, we have:

750w + wt = t.

We can solve this equation to find the value of w, which represents the speed of the river flow.