/*! 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 37 A "moving sidewalk" in an airpor... [FREE SOLUTION] | 91Ó°ÊÓ

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A "moving sidewalk" in an airport terminal building moves at 1.0 \(\mathrm{m} / \mathrm{s}\) and is 35.0 \(\mathrm{m}\) long. If a woman steps on at one end and walks at 1.5 \(\mathrm{m} / \mathrm{s}\) relative to the moving sidewalk, how much time does she require to reach the opposite end if she walks (a) in the same direction the sidewalk is moving? (b) In the opposite direction?

Short Answer

Expert verified
(a) 14.0 seconds in the same direction; (b) 70.0 seconds in the opposite direction.

Step by step solution

01

Understanding the Problem

We have a moving sidewalk that moves at a speed of 1.0 m/s and is 35.0 m long. A woman walks on this sidewalk with a speed of 1.5 m/s; her speed is relative to the sidewalk. We need to calculate the time it takes for her to cover the entire sidewalk length when walking (a) in the same direction as the sidewalk's movement and (b) in the opposite direction.
02

Calculate Speed in the Same Direction

When the woman walks in the same direction as the sidewalk, her total speed relative to the ground is the sum of her walking speed and the sidewalk speed. Therefore, the total speed is 1.5 m/s (woman's speed relative to sidewalk) + 1.0 m/s (sidewalk speed) = 2.5 m/s.
03

Determine Time for Same Direction

The time needed to cover a certain distance is calculated using the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \). Using the combined speed from the previous step, the time is \( \frac{35.0 \, \text{m}}{2.5 \, \text{m/s}} = 14.0 \, \text{seconds} \).
04

Calculate Speed in the Opposite Direction

When the woman walks in the opposite direction to the sidewalk's movement, her speed relative to the ground is the sidewalk's speed subtracted from her walking speed. Therefore, the speed is 1.5 m/s (woman's speed relative to sidewalk) - 1.0 m/s (sidewalk speed) = 0.5 m/s.
05

Determine Time for Opposite Direction

Using the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \), the time it takes is \( \frac{35.0 \, \text{m}}{0.5 \, \text{m/s}} = 70.0 \, \text{seconds} \).

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Key Concepts

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

motion in one dimension
In physics, motion in one dimension refers to the analysis of an object's movement along a straight line. This can either be in a forward or backward direction, such as moving along a sidewalk or a runway. When an object travels in only one direction, it simplifies the understanding of its movement since we don't have to consider any other axes like vertical or lateral directions.
The problem of the woman on the moving sidewalk is an excellent example of motion in one dimension. She moves either along with or against the motion of the conveyor belt which represents different scenarios of linear movement. The concept of motion in one dimension is fundamental in physics because it provides a base to understand more complex motion in two and three dimensions.
When analyzing this type of motion, it is essential to identify the reference point or frame of reference, which in this case is the starting point on the moving sidewalk. This helps us to observe and calculate changes in her position over time.
speed and velocity calculations
Speed and velocity are key concepts in understanding motion. Speed is defined as the distance traveled over a given time and is a scalar quantity meaning it only has magnitude but no direction. On the other hand, velocity is a vector quantity; it includes both magnitude and direction, providing a more detailed description of motion.
In our scenario, we calculate both these quantities to understand how fast the woman moves relative to the moving sidewalk and the ground. When she walks in the same direction as the sidewalk's motion, her total speed relative to the ground becomes 1.0 m/s (speed of the sidewalk) plus 1.5 m/s (her speed relative to the sidewalk). Therefore, her velocity in this direction is 2.5 m/s. This demonstrates how velocity can be influenced by direction and reference points.
Conversely, when walking in the opposite direction, her speed relative to the ground is reduced to 0.5 m/s because the sidewalk's speed (1.0 m/s) opposes her own speed. This inverse calculation helps to understand negative velocities, where motion occurs opposite to the chosen direction of reference.
problem-solving techniques in physics
Approaching problems in physics can be challenging, but using a systematic strategy can make the process easier. Breaking down the problem into smaller steps, understanding known data, and applying relevant physics concepts significantly enhance problem-solving efficiency.
First, interpret the problem by identifying what's given and what needs to be found; like the woman's speed relative to the sidewalk and ground. Setting up equations, like the one for time \(\text{time} = \frac{\text{distance}}{\text{speed}}\), then allows for calculating unknown variables, such as the total time of travel.
Applying these techniques in our example, once you understand the scenario in detail, consider both the additively combined and subtracted speeds for the two walking directions. This method not only provides clarity but also allows you to double-check your solutions by verifying if the dimensions and units align correctly. These insights into organized problem-solving can be valuable for tackling a wide range of physics problems.

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Most popular questions from this chapter

A man stands on the roof of a 15.0 -m-tall building and throws a rock with a velocity of magnitude 30.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. You can ignore air resistance. Calculate (a) the maximum height above the roof reached by the rock; \((b)\) the magnitude of the velocity of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t\) \(v_{x}-t_{2}\) and \(v_{y}-t\) graphs for the motion.

The radius of the earth's orbit around the sun (assumed to be circular) is \(1.50 \times 10^{8} \mathrm{km},\) and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in \(\mathrm{m} / \mathrm{s} ?\) (b) What is the radial acceleration of the earth toward the sun, in \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius \(=5.79 \times 10^{7} \mathrm{km}\) , orbital period \(=88.0\) days.

A rock is thrown from the roof of a building with a velocity \(v_{0}\) at an angle of \(\alpha_{0}\) from the horizontal. The building has height \(h\) You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground, and show that this speed is independent of \(\alpha_{0}\) -

Kicking a Field Goal. In U.S. 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 \(\mathrm{ft}\) horzontally from the bar (Fig. 3.48\() .\) Football regulations are stated in English units, but convert 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 to just clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\) .

A squirrel has \(x-\) and \(y\) -coordinates \((1.1 \mathrm{m}, 3.4 \mathrm{m})\) at time \(t_{1}=0\) and coordinates \((5.3 \mathrm{m},-0.5 \mathrm{m})\) at time \(t_{2}=3.0 \mathrm{s} .\) For this time interval, find (a) the components of the average velocity, and \((\mathrm{b})\) the magnitude and direction of the average velocity.

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