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Chapter 3: Problem 80

Two students are canoeing on a river. While heading upstream, they accidentally drop an empty bottle overboard. They then continue paddling for 60 minutes, reaching a point \(2.0 \mathrm{~km}\) farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they turn around and head downstream. They catch up with and retrieve the bottle (which has been moving along with the current) \(5.0 \mathrm{~km}\) downstream from the turnaround point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

Short Answer

Expert verified
The river is flowing at a speed of 2.5 km/hr and the canoe speed in a still lake for the same paddling effort would be 4.5 km/hr

Step by step solution

01

Analyze the scenario and set up the equations

First, we vision this as two separate events: the boat going upstream and the boat going downstream to catch the bottle. So, \(speed_{upstream} = speed_{stilllake} - speed_{river}\) and \(speed_{downstream} = speed_{stilllake} + speed_{river}\). We know from the problem that the distance covered upstream in 60 minutes (1 hour) is 2 km, so we get \(speed_{upstream}= \frac{2 \mathrm{~km}}{1 \mathrm{~hr}}\). And the distance covered downstream is 5 km more than this, so \(speed_{downstream}=\frac{7 \mathrm{~km}}{1 \mathrm{~hr}}\).
02

Solve Equations

From Step 1, we now have two equations that we will solve simultaneously. Adding the two equations, we get \(2speed_{stilllake} = speed_{upstream} + speed_{downstream}\). Substituting the known speeds, we get \(2speed_{stilllake} = \frac{2 \mathrm{~km}}{1 \mathrm{~hr}} +\frac{7 \mathrm{~km}}{1 \mathrm{~hr}}\). Solving a bit we get \(speed_{stilllake} = 4.5 \mathrm{~km/hr}\). Subtracting the downstream from the upstream equation we get \(2speed_{river}= speed_{downstream}- speed_{upstream} \rightarrow speed_{river}= \frac{7-2}{2} \mathrm{~km/hr} = 2.5 \mathrm{~km/hr}\).
03

Answer the questions

The solution to (a) How fast is the river flowing? is 2.5 km/hr. And for (b) What would the canoe speed in a still lake be for the same paddling effort? is 4.5 km/hr.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

River Current Speed
Understanding the concept of river current speed is essential to analyzing motion in water bodies. River current speed refers to how fast the river itself is flowing in a particular direction. In our exercise, the river's pace affects the canoe's movement both upstream and downstream. This flow is a constant in our problem.

The river current speed offsets the canoe's paddling speed when moving upstream, slowing it down. Conversely, it boosts the canoe's speed when heading downstream.

Think of the river current speed as a conveyor belt that carries everything in its path at a steady rate:
  • Upstream, you must paddle against it, effectively reducing your speed.
  • Downstream, it adds to your paddling speed, increasing your velocity.
In the given exercise, the accurate calculation reveals that the river is flowing at 2.5 km/hr.
Canoe Speed Calculation
When trying to compute the canoe's speed, it is simplified by breaking the journey into upstream and downstream segments. Canoe speed in still water isn't influenced by the river's flow.

To find this value, we use two equations, noting the canoe's behavior against and with the river current:
  • Upstream speed is the result of subtracting the river speed from the canoe's speed in still water.
  • Downstream speed is found by adding river speed to the canoe's still water speed.
The given exercise shows the process of deriving these speeds using distance-time analysis.

With the calculations shown, the canoe's pace in still water is 4.5 km/hr, found by averaging the contrasting effects of current on the journey
Distance-Time Analysis
Distance-time analysis is a key tool to resolve the canoe and river problem. By looking at how far the canoe travels over a period, we relate speed to time and distance to solve the puzzle.

In our scenario, the students travel 2 km upstream in one hour before realizing the bottle is missing. Afterwards, they move 7 km downstream over the same time span to catch the bottle.

This analysis is important in understanding the movement dynamics:
  • Separate the journey into segments: upstream and downstream.
  • Measure the distances and calculate speeds accordingly.
By comprehensively analyzing time and distance, we effectively determine the speed of the river current and the canoe's speed in still water, marrying calculations to the physical scenario.

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Most popular questions from this chapter

A railroad flatcar is traveling to the right at a speed of \(13.0 \mathrm{~m} / \mathrm{s}\) relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. E3.36). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) \(18.0 \mathrm{~m} / \mathrm{s}\) to the right? (b) \(3.0 \mathrm{~m} / \mathrm{s}\) to the left? (c) zero?

In Canadian football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is \(10.0 \mathrm{ft}\) above the ground, and the ball is kicked from ground level, \(36.0 \mathrm{ft}\) horizontally from the bar (Fig. \(\mathbf{P 3 . 6 0}\) ). Football regulations are stated in English units, but convert them to SI units for this problem. (a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at \(45.0^{\circ}\) above the horizontal, what must its initial speed be if it is just to clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\).

The position of a dragonfly that is flying parallel to the ground is given as a function of time by \(\vec{r}=[2.90 \mathrm{~m}+\) \(\left.\left(0.0900 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\imath}-\left(0.0150 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} \hat{\jmath} .\) (a) At what value of \(t\) does the velocity vector of the dragonfly make an angle of \(30.0^{\circ}\) clockwise from the \(+x\) -axis? (b) At the calculated in part (a), what are the magnitude and direction of the dragonfly's acceleration vector?

Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about \(100 \mathrm{~km} / \mathrm{h}\). If one goose is flying at \(100 \mathrm{~km} / \mathrm{h}\) relative to the air but a \(40 \mathrm{~km} / \mathrm{h}\) wind is blowing from west to east, (a) at what angle relative to the north-south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of \(500 \mathrm{~km}\) from north to south? (Note: Even on cloudy nights, many birds can navigate by using the earth's magnetic field to fix the northsouth direction.)

A baseball thrown at an angle of \(60.0^{\circ}\) above the horizontal strikes a building \(18.0 \mathrm{~m}\) away at a point \(8.00 \mathrm{~m}\) above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.
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