/*! 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 42 A river has a steady speed of \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A river has a steady speed of \(v_{1}\) A student swims upstream a distance \(d\) and back to the starting point. (a) If the student can swim at a speed of \(v\) in still water, how much time \(t_{\text {op }}\) does it take the student to swim upstream a distance \(d ?\) Express the answer in terms of \(d\). \(v_{\text {, }}\) and \(v_{s}\). (b) Using the same variables, how much time \(t_{\text {doun }}\) does it take to swim back downstream to the starting point? (c) Sum the answers found in parts (a) and (b) and show that the time \(t_{a}\) required for the whole trip can be written as $$t_{\mathrm{a}}=\frac{2 d / v}{\mathrm{I}-\mathrm{v}_{1}^{2} / v^{2}}$$ (d) How much time \(t_{b}\) does the trip take in still water? (e) Which is larger, \(l_{a}\) or \(t_{b}^{2}\) Is it always larger?

Short Answer

Expert verified
The calculated times are \(t_{up} = \frac{d}{v - v_{1}}\), \(t_{down} = \frac{d}{v + v_{1}}\), \(t_a = \frac{2d/v}{1 - (v_{1}^2 / v^2)}\) and \(t_b = \frac{2d}{v}\). It can be concluded that under most conditions \(t_{a} > t_{b}\)

Step by step solution

01

Calculate Time to Swim Upstream

To swim upstream a distance 'd', the student will swim at a speed of \(v - v_{1}\) because the current is against their swimming. Thus, the time taken can be calculated by the general formula for time which is Distance/Speed. It will therefore be \(t_{up} = \frac{d}{v - v_{1}}\)
02

Calculate Time to Swim Downstream

To swim downstream a distance 'd', the student will have the advantage of the river's current. Thus, they will effectively swim at a speed of \(v + v_{1}\). Using the same formula used in Step 1, the time taken to go downstream will be \(t_{down} = \frac{d}{v + v_{1}}\)
03

Sum the two times to get the total time of the trip

The total time of the trip is the summation of the time taken to go upstream and that taken to come downstream. Therefore, \(t_a = t_{up} + t_{down} = \frac{d}{v - v_{1}} + \frac{d}{v + v_{1}}\). By finding the common denominator and simplifying the fraction, we get \(t_{a} = \frac{2d/v}{1 - (v_{1}^2 / v^2)}\)
04

Calculate Time for Trip in Still Water

In still water where there's no current, the student swims at a speed 'v' up and down the river. This time it will be calculated as \(t_b = \frac{2d}{v}\)
05

Comparison of \(t_{a}\) and \(t_{b}^2\)

Notice that \(t_{b}^2\) expands to \(\frac{4d^2}{v^2}\), while the formula for \(t_{a}\) has a denominator of \(1 - (v_{1}^2 / v^2)\), The square roots of both times are compared we can deduce that \(t_{a} \geq t_{b}\) for all non-zero speeds. This is because (1) the swimmer will always be slower when the current is against them, and (2) the effect of the current assisting the swimmer when returning downstream does not completely compensate for the slower rate upstream. The only time \(t_{a} = t_{b}\) is when the current speed is 0 and thus \(t_{a} > t_{b}\) in all other cases.

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Key Concepts

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

Relative Velocity
Understanding relative velocity is crucial when analyzing motion in mediums like water or air. In the context of the problem, the concept of relative velocity comes into play when the student swims upstream and downstream in a river. As water flows in a certain direction, it impacts the speed at which the student is effectively swimming.

Relative velocity is the velocity of an object as observed from a particular frame of reference, which in this case is the flowing river. When the student swims upstream, the river's current opposes their swimming speed, reducing their effective swimming speed. Conversely, when swimming downstream, the river's current aids their swimming speed, increasing their effective speed. Hence, the student's effective velocity can be expressed as:

  • Upstream: effective velocity = swimming speed - river current speed (v_{effective} = v - v_{1})
  • Downstream: effective velocity = swimming speed + river current speed (v_{effective} = v + v_{1})
This understanding helps calculate travel times accurately across differing scenarios.
Time Calculation in Physics
Calculating time in physics often depends on the relationship between distance, speed, and time. This relationship is summed up by the formula: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \].

In this exercise, the journey is broken into two parts: upstream and downstream swimming. The time taken for each part can be calculated by applying this formula with the respective relative velocities.

  • For the upstream journey, time is calculated as: \[ t_{up} = \frac{d}{v - v_{1}} \]
  • For the downstream journey, the calculation is: \[ t_{down} = \frac{d}{v + v_{1}} \]
By understanding this core principle, students can decompose and analyze each segment of the trip effectively by employing relative speeds specific to each context.
Physics Problem Solving
Problem solving in physics often requires a structured approach to dissect complex scenarios into manageable parts. This problem demands an understanding of variable interactions such as velocity, distance, and time, alongside their possible effects on tasks like swimming across a river.

Here's a simplified approach to resolving such exercises:
  • Identify knowns and unknowns: Start by defining variables and their known values. In this exercise, knowns include swimming speed and river current speed.
  • Break down scenarios: As shown, divide the problem into distinct components—upstream travel and downstream return.
  • Apply appropriate formulas: Use physics equations such as \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] with context-specific velocities for each scenario.
  • Sum results and simplify: After individual calculations, combine results to find the solution, simplifying where possible for clarity and accuracy.
This systematic approach makes tackling even complex physics problems much more manageable.
Kinematics in Water
Kinematics in water involves analyzing motion considering external factors like water currents. When an object moves through water, its movement is influenced by the water's motion, necessitating adjustments in calculations of speed and time.

As seen in our swimmer's problem, understanding these dynamics helps in calculating not just direct velocity but relative motion compared to moving mediums. In this setup:
  • Understand the medium: Recognize how water current affects speed and direction.
  • Determine effective speed: Calculate how the current adds to or subtracts from the swimmer's speed, impacting overall travel time.
By integrating these elements, students can explore more complex kinematic scenarios in aquatic environments, enhancing their comprehension of motion through different media, and grasping the reality of opposing forces within water dynamics.

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Most popular questions from this chapter

A chinook (king) salmon (genus Oncorhynchus) can jump out of water with a speed of \(6.26 \mathrm{~m} / \mathrm{s}\). (See Problem \(4.9\), page 111 for an investigation of how the fish can leave the water at a higher speed than it can swim underwater.) If the salmon is in a stream with water speed equal to \(1.50 \mathrm{~m} / \mathrm{s}\), how high in the air can the fish jump if it leaves the water traveling vertically upwards relative to the Earth?

A jogger runs \(100 \mathrm{~m}\) due west, then changes direction for the second leg of the run. At the end of the run, she is 175 \(\mathrm{m}\) away from the starting point at an angle of \(15.0^{\circ}\) north of west. What were the direction and length of her second displacement? Use graphical techniques.

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A small map shows Atlanta to be 730 miles in a direction \(5^{\circ}\) north of east from Dallas. The same map shows that Chicago is 560 miles in a dircction \(21^{\text {n }}\) west of north from Atlanta. Assume a flat Earth and use the given information to find the displacement from Dallas to Chicago.

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