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Chapter 18: Problem 9

Starting \(3.5 \mathrm{m}\) from a department store mirror, Suzanne walks toward the mirror at \(1.5 \mathrm{m} / \mathrm{s}\) for \(2.0 \mathrm{s}\). How far is Suzanne from her image in the mirror after \(2.0 \mathrm{s}\) ?

Short Answer

Expert verified
The distance from Suzanne to her image in the mirror after \(2.0 \mathrm{s}\) is \(1.0 \mathrm{m}\).

Step by step solution

01

Calculate the distance moved

Suzanne's speed is \(1.5 \mathrm{m} / \mathrm{s}\), so in \(2.0 \mathrm{s}\) she will have walked \(1.5 \mathrm{m}/\mathrm{s} \times 2.0 \mathrm{s} = 3.0 \mathrm{m}\).
02

Calculate the final distance from the mirror

She started \(3.5 \mathrm{m}\) away from the mirror, and moved \(3.0 \mathrm{m}\) closer to it. So she is \(3.5 \mathrm{m} - 3.0 \mathrm{m} = 0.5 \mathrm{m}\) from the mirror after \(2.0 \mathrm{s}\).
03

Calculate the total distance to her image

Suzanne's image in the mirror is the same distance behind the mirror as she is in front of it, so the total distance from Suzanne to her image is \(0.5 \mathrm{m} + 0.5 \mathrm{m} = 1.0 \mathrm{m}\).

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Key Concepts

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

Distance
Distance is a fundamental concept in kinematics, which describes how far an object moves from one point to another. In this exercise, Suzanne begins her walk at a distance of \(3.5 \mathrm{m}\) from a mirror. As she moves closer, the distance changes according to her starting point and the distance she covers. To find the total distance covered during her walk, we multiply her speed by the time taken:
  • Speed: \(1.5 \mathrm{m/s}\)
  • Time: \(2.0 \mathrm{s}\)
  • Distance moved: \(1.5 \mathrm{m/s} \times 2.0 \mathrm{s} = 3.0 \mathrm{m}\)
By the end of \(2.0 \mathrm{s}\), the final distance from her initial position to the new one helps us understand how far she has traveled.
Image Formation
Image formation refers to the virtual image that appears behind a mirror when an object is placed in front of it. This principle plays a vital role in calculating Suzanne's distance to her image in the mirror. Mirrors reflect light to create this image, which appears to be the same distance behind the mirror as the object is in front of it. Hence, if Suzanne is \(0.5 \mathrm{m}\) from the mirror post-walk, her image also seems to be \(0.5 \mathrm{m}\) on the opposite side.The entire process of image formation helps us understand this fascinating reflection principle that mirrors employ, and why Suzanne sees her image as if it's on the other side of the mirror.
Speed
Speed is a measure of how quickly an object moves. It is defined as the distance covered per unit time. In kinematics, understanding speed is crucial to computing how far an object will travel over a certain period.For Suzanne, her speed is \(1.5 \mathrm{m/s}\), meaning she covers \(1.5 \mathrm{m}\) every second. To determine how far she walks toward the mirror, we multiply her speed by the duration of her movement:
  • Speed formula: Distance = Speed 脳 Time
  • Example: \(1.5 \mathrm{m/s} \times 2.0 \mathrm{s} = 3.0 \mathrm{m}\)
This calculation represents how kinematics employs simple equations to break down real-world movements, showing us how far Suzanne progresses in her 2-second walk.
Mirrors
Mirrors are reflective surfaces that create images of objects placed in front of them. These formed images appear as if they are positioned behind the mirror due to reflection principles.In Suzanne's case, the mirror plays a significant role. After she walks closer to it, the distance to her image is calculated using the property of mirrors where the image is as far inside the mirror as the object is outside:
  • Distance from Suzanne to mirror after walking: \(0.5 \mathrm{m}\)
  • Total distance from Suzanne to her image: \(0.5 \mathrm{m} + 0.5 \mathrm{m} = 1.0 \mathrm{m}\)
The functionality of mirrors in optics demonstrates not only how images are perceived but also supports the calculations necessary in kinematic problems involving reflections.

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Most popular questions from this chapter

A slide projector needs to create a 98 -cm-high image of a 2.0-cm-tall slide. The screen is \(300 \mathrm{cm}\) from the slide. a. What focal length does the lens need? Assume that it is a thin lens. b. How far should you place the lens from the slide?

To a fish, the \(4.00-\mathrm{mm}\) -thick aquarium walls appear only \(3.50 \mathrm{mm}\) thick. What is the index of refraction of the walls?

The glass core of an optical fiber has index of refraction 1.60. The index of refraction of the cladding is \(1.48 .\) What is the maximum angle between a light ray and the wall of the core if the ray is to remain inside the core?

The sun is \(150,000,000 \mathrm{km}\) from earth; its diameter is \(1,400,000 \mathrm{km}\) For a science project on solar For a science project on solar power, a student uses a 24 -cm-diameter converging mirror with a focal length of \(45 \mathrm{cm}\) to focus sunlight onto an object. This casts an image of the sun on the object. For the most intense heat, the image of the sun should be in focus. a. Where should the object be placed? b. What is the diameter of the image? c. The intensity of the incoming sunlight is \(1050 \mathrm{W} / \mathrm{m}^{2} .\) What is the total power of the light captured by the mirror? d. What is the intensity of sunlight in the projected image? Assume that all of the light captured by the mirror is focused into the image.

Calculate the image position and height. A 3.0 -cm-tall object is \(15 \mathrm{cm}\) in front of a converging mirror that has a \(25 \mathrm{cm}\) focal length.
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