/*! 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 An airport terminal has a moving... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An airport terminal has a moving sidewalk to speed passengers through a long corridor. Larry does not use the moving sidewalk; he takes \(150 \mathrm{~s}\) to walk through the corridor. Curly, who simply stands on the moving sidewalk, covers the same distance in 70 s. Moe boards the sidewalk and walks along it. How long does Moe take to move through the corridor? Assume that Larry and Moe walk at the same speed.

Short Answer

Expert verified
47.73 seconds approximately.

Step by step solution

01

Define Variables and Known Information

Let's denote Larry's walking speed as \( v_L \), the speed of the moving sidewalk as \( v_m \), and the corridor distance as \( d \). From the problem, we know:- Time taken by Larry to walk through the corridor: 150 s.- Time taken by Curly on the moving sidewalk: 70 s.The walking speed of Larry: \( v_L = \frac{d}{150} \).The speed of the moving sidewalk: \( v_m = \frac{d}{70} \).
02

Determine Moe's Total Speed

Since Moe walks on the moving sidewalk, his speed is the sum of his walking speed and the moving sidewalk's speed. Therefore, Moe's speed \( v_M \) can be expressed as:\[ v_M = v_L + v_m = \frac{d}{150} + \frac{d}{70} \]
03

Simplify and Solve for Moe's Time

To find how long Moe takes, we need to determine \( t_M \) using Moe's speed:Since \( v_M = \frac{d}{t_M} \), we can set:\[ \frac{d}{t_M} = \frac{d}{150} + \frac{d}{70} \]Cancel the distance \( d \) from both sides:\[ \frac{1}{t_M} = \frac{1}{150} + \frac{1}{70} \]To solve, we find the least common denominator (LCD) of 150 and 70, which is 1050:\[ \frac{1}{t_M} = \frac{7}{1050} + \frac{15}{1050} = \frac{22}{1050} \]So,\[ t_M = \frac{1050}{22} \approx 47.73 \text{ seconds} \]
04

Conclusion

Moe, walking on the moving sidewalk, takes approximately 47.73 seconds to travel through the corridor.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Speed and Velocity
Speed and velocity are fundamental concepts in understanding motion problems related to time, distance, and movement, such as the one given in this exercise. Speed is a scalar quantity that refers to how fast an object is moving, without consideration of the object's direction of travel. It is generally calculated using the formula:
  • Speed = Distance / Time
Velocity, on the other hand, is a vector quantity. This means it not only considers the speed of an object but also the direction in which the object is moving. In most introductory motion problems, when direction isn’t specifically mentioned, speed and velocity can be calculated similarly. However, understanding the distinction ensures accurate interpretations in physics problems where direction plays a key role.
This problem shows how to use speed to determine the walking time of different individuals over the same distance, yet under different conditions such as walking normally, using a moving sidewalk, or combining both.
Time and Distance
Time and distance are closely linked in problems involving movement. The time it takes to cover a distance can be altered by changing the speed of travel. For instance, in the scenario described, Larry takes 150 seconds to cover the corridor by walking, indicating that his speed is relatively slower compared to the combined speed of walking on a moving sidewalk.
To solve the problem, we identify the relationship:
  • Larry’s walking speed: \( v_L = \frac{d}{150} \)
  • The moving sidewalk’s speed: \( v_m = \frac{d}{70} \)
These equations reflect how effectively time and distance interact in motion problems.
When you calculate time by rearranging the formula to Time = Distance / Speed, you see how changes in either distance or speed will affect time. Here, the strategic combination of walking speed and sidewalk speed allows Moe to traverse the same distance more quickly than Larry.
Relative Motion
Relative motion is crucial when analyzing how different elements in a system contribute to overall speed. In this problem, Moe's motion on the moving sidewalk combines both his walking speed and the sidewalk's speed.
Relative motion considers how one object's movement compares or adds to another's. In practical terms, Moe's total speed becomes a sum, allowing calculations to involve multiple motion influences:
  • Moe’s total speed: \( v_M = v_L + v_m \)
By adding the speeds, you account for both his individual walking speed and the moving sidewalk’s speed.
Then, to find the time Moe takes, rearrange the formula as shown:
  • \( \frac{1}{t_M} = \frac{1}{150} + \frac{1}{70} \)
This equation uses the concept of inverse addition for relative speeds, leading to the calculation of endpoint time, which represents how relative motion can simplify time in dynamic systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A certain airplane has a speed of \(290.0 \mathrm{~km} / \mathrm{h}\) and is diving at an angle of \(\theta=30.0^{\circ}\) below the horizontal when the pilot releases a radar decoy (Fig. \(4-33\) ) . The horizontal distance between the release point and the point where the decoy strikes the ground is \(d=700 \mathrm{~m} .\) (a) How long is the decoy in the air? (b) How high was the release point?

A transcontinental flight of \(4350 \mathrm{~km}\) is scheduled to take 50 min longer westward than eastward. The airspeed of the airplane is \(966 \mathrm{~km} / \mathrm{h},\) and the jet stream it will fly through is presumed to move due east. What is the assumed speed of the jet stream?

Suppose that a shot putter can put a shot at the world-class speed \(v_{0}=15.00 \mathrm{~m} / \mathrm{s}\) and at a height of \(2.160 \mathrm{~m} .\) What horizontal distance would the shot travel if the launch angle \(\theta_{0}\) is (a) \(45.00^{\circ}\) and (b) \(42.00^{\circ}\) ? The answers indicate that the angle of \(45^{\circ},\) which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

Two seconds after being projected from ground level, a projectile is displaced \(40 \mathrm{~m}\) horizontally and \(53 \mathrm{~m}\) vertically above its launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

A person walks up a stalled 15-m-long escalator in \(90 \mathrm{~s}\). When standing on the same escalator, now moving, the person is carried up in \(60 \mathrm{~s}\). How much time would it take that person to walk up the moving escalator? Does the answer depend on the length of the escalator?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.