/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Two swimmers, Alan and Beth, sta... [FREE SOLUTION] | 91Ó°ÊÓ

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Two swimmers, Alan and Beth, start together at the same point on the bank of a wide stream that flows with a speed v. Both move at the same speed \(c(c>v),\) relative to the water. Alan swims downstream a distance \(L\) and then \(\mathrm{up}^{-}\) stream the same distance. Beth swims so that her motion relative to the Earth is perpendicular to the banks of the stream. She swims the distance \(L\) and then back the same distance, so that both swimmers return to the starting point. Which swimmer returns first? (Note: First guess the answer.)

Short Answer

Expert verified
The swimmer who returns to the starting point first is the one whose total time \(T\) is smaller, either Alan if \(T_{A} < T_{B}\), or Beth if \(T_{B} < T_{A}\). Without specific values for \(v\), \(c\), and \(L\), it's not possible to definitively answer which one returns first.

Step by step solution

01

Calculate time taken by Alan

Alan swims downstream and then upstream a distance \(L\). The speed of Alan relative to ground while going downstream is \(v+c\) and while going upstream is \(c-v\). Thus, the total time taken by Alan, \(T_{A}\) can be calculated using the formula \[T_{A} = \frac{L}{v+c} + \frac{L}{c-v}\]
02

Calculate time taken by Beth

Beth swims perpendicular to the stream, so her speed relative to ground is \(\sqrt{c^2 - v^2}\) (due to the Pythagorean theorem). As she swims the same distance \(L\) twice, the total time taken by Beth, \(T_{B}\) can be calculated using the formula \[T_{B} = 2* \frac{L}{\sqrt{c^2 - v^2}}\]
03

Compare the times

By comparing the total time taken by Alan and Beth, \(T_{A}\) and \(T_{B}\), it can be determined who returns to the starting point first.

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Key Concepts

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

Relative Motion in Physics
Understanding relative motion is central to solving problems like the swimmer crossing a flowing stream. The crucial concept here is that the velocity of an object must always be considered relative to a particular frame of reference. When Alan swims down and upstream, his motion is affected by the velocity of the stream, v. In contrast, Beth's swimming direction is perpendicular to the stream, and thus, her situation requires a different analytical approach, using the Pythagorean theorem.

To visualize relative motion, imagine standing on a moving walkway at an airport. Your speed relative to the walkway might be constant, but your speed relative to the ground changes based on whether the walkway is moving with you or against you. Similarly, Alan's effective speed varies when swimming with and against the stream, and this change must be accounted for when calculating his time to travel a certain distance.
Pythagorean Theorem Application
Beth's challenge in swimming across the stream can be pictured as a right-angled triangle where her speed c relative to the water forms the hypotenuse, and the water's velocity v forms one of the sides. The side representing the swimmer's effective velocity relative to the riverbank is what we are solving for. Here's where the Pythagorean theorem comes into play: a fundamental formula describing the relationship between the lengths of the sides of a right-angled triangle.

The theorem is stated as \(a^2 + b^2 = c^2\), with \(a\) and \(b\) representing the sides adjacent to the right angle, and \(c\) being the hypotenuse. For Beth, the effective speed relative to the ground, which is the side adjacent to the river flow vector, can be found by rearranging the theorem to solve for \(b\), yielding \( b = \sqrt{c^2 - v^2} \). This effective velocity is pertinent for calculating how long it takes for Beth to swim to a certain point and back.
Time Calculation with Velocity
When solving motion problems, the time calculation is often a straightforward application of the basic formula for time: \(t = \frac{d}{v}\), where \(t\) is time, \(d\) is distance, and \(v\) is velocity. The intricacy arises when dealing with varying velocities, as seen with Alan's swimming scenario. We must calculate his time for each leg of his swim separately, due to the differing velocities downstream and upstream. For Beth, the constant effective velocity allows for a simpler time calculation.

The swimmer's problems often test the understanding of how to manipulate these fundamental equations to account for the relative velocities resulting from external motions, such as the flow of water. Grasping the concepts of relative velocity, the Pythagorean theorem in motion, and the calculation of time with velocity equips students with tools to decipher problems that extend well beyond the riverbank.

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Most popular questions from this chapter

The water in a river flows uniformly at a constant speed of \(2.50 \mathrm{m} / \mathrm{s}\) between parallel banks \(80.0 \mathrm{m}\) apart. You are to deliver a package directly across the river, but you can swim only at \(1.50 \mathrm{m} / \mathrm{s} .\) (a) If you choose to minimize the time you spend in the water, in what direction should you head? (b) How far downstream will you be carried? (c) What If? If you choose to minimize the distance downstream that the river carries you, in what direction should you head? (d) How far downstream will you be carried?

A hawk is flying horizontally at \(10.0 \mathrm{m} / \mathrm{s}\) in a straight line, \(200 \mathrm{m}\) above the ground. A mouse it has been carrying struggles free from its grasp. The hawk continues on its path at the same speed for 2.00 seconds before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse \(3.00 \mathrm{m}\) above the ground. (a) Assuming no air resistance, find the diving speed of the hawk. (b) What angle did the hawk make with the horizontal during its descent? (c) For how long did the mouse "enjoy" free fall?

Your grandfather is copilot of a bomber, flying horizontally over level terrain, with a speed of 275 m/s relative to the ground, at an altitude of 3000 m. (a) The bombardier releases one bomb. How far will it travel horizontally between its release and its impact on the ground? Neglect the effects of air resistance. (b) Firing from the people on the ground suddenly incapacitates the bombardier before he can call, "Bombs away!" Consequently, the pilot maintains the plane's original course, altitude, and speed through a storm of flak. Where will the plane be when the bomb hits the ground? (c) The plane has a telescopic bomb sight set so that the bomb hits the target seen in the sight at the time of release. At what angle from the vertical was the bomb sight set?

Barry Bonds hits a home run so that the baseball just clears the top row of bleachers, \(21.0 \mathrm{m}\) high, located \(130 \mathrm{m}\) from home plate. The ball is hit at an angle of \(35.0^{\circ}\) to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time at which the ball reaches the cheap seats, and (c) the velocity components and the speed of the ball when it passes over the top row. Assume the ball is hit at a height of \(1.00 \mathrm{m}\) above the ground.

Young David who slew Goliath experimented with slings before tackling the giant. He found that he could revolve a sling of length \(0.600 \mathrm{m}\) at the rate of 8.00 rev/s. If he increased the length to \(0.900 \mathrm{m},\) he could revolve the sling only 6.00 times per second. (a) Which rate of rotation gives the greater speed for the stone at the end of the sling? (b) What is the centripetal acceleration of the stone at 8.00 rev/s? (c) What is the centripetal acceleration at \(6.00 \mathrm{rev} / \mathrm{s} ?\)

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