/*! 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 118 The wings on a stonefly do not f... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 118

The wings on a stonefly do not flap, and thus the insect cannot fly. However, when the insect is on a water surface, it can sail across the surface by lifting its wings into a breeze. Suppose that you time stoneflies as they move at constant speed along a straight path of a certain length. On average, the trips each take \(7.1 \mathrm{~s}\) with the wings set as sails and \(25.0 \mathrm{~s}\) with the wings tucked in. (a) What is the ratio of the sailing speed \(v_{s}\) to the nonsailing speed \(v_{n s} ?(\mathrm{~b})\) In terms of \(v_{s}\), what is the difference in the times the insects take to travel the first \(2.0 \mathrm{~m}\) along the path with and without sailing?

Short Answer

Expert verified
The sailing speed is 3.52 times faster. The time difference for 2 m is approximately \( \frac{5}{v_s} \) seconds.

Step by step solution

01

Understand the Problem

We need to find the ratio of the sailing speed to the nonsailing speed, and then determine the time difference for a 2 m journey with and without sailing. We're given the times for each mode of travel along a certain path.
02

Define Variables and Concepts

Let the length of the path be \( L \), the sailing speed be \( v_s \), and the nonsailing speed be \( v_{ns} \). We have \( t_s = 7.1 \) s and \( t_{ns} = 25.0 \) s. The relationship for constant speed is \( v = \frac{L}{t} \).
03

Calculate Sailing and Nonsailing Speeds

Since \( v_s = \frac{L}{t_s} \) and \( v_{ns} = \frac{L}{t_{ns}} \), we can express the ratio \( \frac{v_s}{v_{ns}} = \frac{t_{ns}}{t_s} \). Substitute the given times: \( \frac{v_s}{v_{ns}} = \frac{25.0}{7.1} \).
04

Calculate Speed Ratio

Perform the division: \( \frac{25.0}{7.1} \approx 3.52 \). So, \( \frac{v_s}{v_{ns}} \approx 3.52 \).
05

Determine Time with Sailing for 2 m

With sailing, \( t_s = \frac{2.0}{v_s} \). Substitute \( v_s = \frac{L}{7.1} \) leading to \( t_s = \frac{2.0 \times 7.1}{L} \).
06

Determine Time without Sailing for 2 m

Without sailing, \( t_{ns} = \frac{2.0}{v_{ns}} \). Substitute \( v_{ns} = \frac{L}{25.0} \) leading to \( t_{ns} = \frac{2.0 \times 25.0}{L} \).
07

Calculate Time Difference for 2 m

The time difference \( \Delta t = t_{ns} - t_s = \frac{2.0 \times 25.0}{L} - \frac{2.0 \times 7.1}{L} = \frac{2.0 (25.0 - 7.1)}{L} = \frac{2.0 \times 17.9}{L} \).
08

Simplify Time Difference in terms of \(v_s\)

Since \( L = v_s \times 7.1 \), substitute it into the previous expression: \( \Delta t = 2.0 \times \frac{17.9}{v_s \times 7.1} = \frac{35.8}{7.1 v_s} \approx \frac{5}{v_s} \).

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause such motions. It focuses on various parameters, such as velocity, acceleration, displacement, and time. In this exercise, kinematics helps us determine how fast and in what time frame the stonefly moves when using its wings as sails versus when it keeps its wings tucked in.
Understanding the fundamental kinematic equation, \( v = \frac{L}{t} \), is crucial here. This equation states that velocity \( v \) is equal to the displacement \( L \) divided by the time \( t \). When studying constant speed motion, as in the case of the stonefly, knowing the time taken for specific distances allows us to calculate the velocity directly. This makes kinematics a powerful tool for problem-solving in physics.
Speed Ratio
When analyzing the motion of an object, the speed ratio is a valuable concept that compares how two different speeds relate to one another. In this particular problem, the exercise asks for the ratio of the stonefly's speed when sailing compared to when it's not sailing.
The speed ratio is calculated using the relationship \( \frac{v_s}{v_{ns}} = \frac{t_{ns}}{t_s} \), where \( v_s \) is the sailing speed and \( v_{ns} \) is the nonsailing speed. Given that the stonefly takes 25.0 seconds to complete a certain path with its wings tucked and 7.1 seconds with its wings set as sails, we see how significantly faster it can move by harnessing the breeze. By calculating the ratio \( \frac{25.0}{7.1} \approx 3.52 \), we understand that the stonefly sails at over three times the speed as when it does not sail.
Constant Speed Motion
Constant speed motion refers to when an object's speed remains uniform throughout its journey. This is applicable in understanding how the stonefly moves across the water. When an object moves at a constant speed, the total distance covered over a period divided by the time taken will yield the same speed at any instant during the motion.
For the stonefly, whether with wings as sails or tucked, the speed remains constant during each respective mode of travel. Using the concept of constant speed motion, we derived the time for a 2 m journey: when sailing, \( t_s = \frac{2.0}{v_s} \), and when not sailing, \( t_{ns} = \frac{2.0}{v_{ns}} \). This allows us to pinpoint and compare specific travel times, enhancing our understanding of motion and speed dynamics.

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Most popular questions from this chapter

A hoodlum throws a stone vertically downward with an initial speed of \(12.0 \mathrm{~m} / \mathrm{s}\) from the roof of a building, \(30.0 \mathrm{~m}\) above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact?

Compute your average velocity in the following two cases: (a) You walk \(73.2 \mathrm{~m}\) at a speed of \(1.22 \mathrm{~m} / \mathrm{s}\) and then run \(73.2 \mathrm{~m}\) at a speed of \(3.05 \mathrm{~m} / \mathrm{s}\) along a straight track. (b) You walk for \(1.00 \mathrm{~min}\) at a speed of \(1.22 \mathrm{~m} / \mathrm{s}\) and then run for \(1.00 \mathrm{~min}\) at \(3.05 \mathrm{~m} / \mathrm{s}\) along a straight track. (c) Graph \(x\) versus \(t\) for both cases and indicate how the average velocity is found on the graph.

To set a speed record in a measured (straight-line) distance \(d\), a race car must be driven first in one direction (in time \(t_{1}\) ) and then in the opposite direction (in time \(t_{2}\) ). (a) To eliminate the effects of the wind and obtain the car's speed \(v_{c}\) in a windless situation, should we find the average of \(d / t_{1}\) and \(d / t_{2}\) (method 1) or should we divide \(d\) by the average of \(t_{1}\) and \(t_{2} ?\) (b) What is the fractional difference in the two methods when a steady wind blows along the car's route and the ratio of the wind speed \(v_{w}\) to the car's speed \(v_{c}\) is \(0.0240 ?\)

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When startled, an armadillo will leap upward. Suppose it rises \(0.544 \mathrm{~m}\) in the first \(0.200 \mathrm{~s}\). (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of \(0.544 \mathrm{~m}\) ? (c) How much higher does it go?
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