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Chapter 36: Problem 16

Now suppose you are in front of a large vertical plane mirror and running toward it at a constant \(5.0 \mathrm{~m} / \mathrm{s}\). How fast will you be approaching your image?

Short Answer

Expert verified
You approach your image at 10.0 m/s.

Step by step solution

01

Understanding the Scenario

When you run toward a mirror, your image appears to also move toward you at the same speed that you are running towards the mirror. Thus, the relative speed between you and your image is the speed at which you are moving plus the speed of the image moving toward you.
02

Determining Relative Speed

You are moving toward the mirror at a speed of 5.0 m/s. Since the mirror reflects your image moving toward you at the same speed, your image is also moving toward the mirror at 5.0 m/s. Therefore, the relative speed between you and your image is: \[ 5.0 ext{ m/s} + 5.0 ext{ m/s} = 10.0 ext{ m/s} \]
03

Calculation

By adding the speed at which you and your image move towards each other due to the reflection, we find that you are approaching your image at a speed of 10.0 m/s.

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Key Concepts

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

Understanding Relative Motion
Relative motion refers to the concept of comparing the velocity of an object with respect to another object. In our exercise, you are running towards a mirror, which is stationary, while your reflection (image) is moving towards you.

When you consider the motion relative to the mirror, both you and your image appear to be running towards each other with an equal speed. The idea here is that their relative velocities add up. Think of it like two cars moving towards each other on a straight road. Each car's speed would contribute to the total speed between them.

Relative motion is a key concept in physics that helps us understand how different frames of reference observe the same event or scenario differently. In this case, the calculation simplifies to merely doubling your speed as you and your image both move at the same rate.
Reflection in Mirrors
Mirrors are fascinating tools that allow us to see virtual images of ourselves. When you stand in front of a large vertical mirror, the light from your body reflects off the mirror back into your eyes, allowing you to see your image.

The image appears to move at the same speed and in the opposite direction of your movement relative to the mirror. This happens due to the laws of reflection, which state that the angle at which light hits the mirror is equal to the angle at which it reflects. Therefore, when you move towards the mirror, your image, as a result of reflection, appears to move towards you.

Understanding the principles of reflection is crucial to grasping why your image's relative motion equals your speed plus the speed of your movement in the original scenario.
Calculating Velocity
Velocity is simply the speed of something in a given direction. When calculating how fast you and your image approach each other, you consider both your speed and the image's speed, derived from reflection.

Let's break it down:
  • Your velocity towards the mirror: 5.0 m/s.
  • Your image's velocity toward you: 5.0 m/s.
Together, these velocities produce a relative velocity of 10.0 m/s. This calculation involves understanding that velocity is a vector quantity, involving both magnitude (speed) and direction. So when both you and your image move towards each other with equal speeds, the magnitudes add up.

Velocity calculations in physics help us determine exact speeds including direction, and are essential for problems involving motion and reflections like the one in our example.

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Most popular questions from this chapter

If you wish to take a photo of yourself as you stand \(3 \mathrm{~m}\) in front of a plane mirror, for what distance should you focus the camera you are holding?

As shown in Fig. \(36-11\), an object \(6 \mathrm{~cm}\) high is located \(30 \mathrm{~cm}\) in front of a convex spherical mirror of radius \(40 \mathrm{~cm} .\) Determine the position and height of its image, \((a)\) by construction and \((b)\) by use of the mirror equation. (a) Choose two convenient rays coming from \(O\) at the top of the object: (1) A ray OA, parallel to the principal axis, is reflected in the direction \(A A^{\prime}\) as if it passed through the principal focus \(F\). (2) A ray OB, directed toward the center of curvature \(C\), is normal to the mirror and is reflected back on itself. The reflected rays, \(A A^{\prime}\) and \(B O\), never meet but appear to originate from a point \(I\) behind the mirror. Then \(I I^{\prime}\) represents the size and position of the image of \(O O^{\prime}\). All images formed by convex mirrors are virtual, erect, and reduced in size, provided the object is in front of the mirror (i.e., a real object). For a convex mirror the radius is positive; here \(R=40\) \(\mathrm{cm} .\) And so $$\begin{array}{l} \text { (b) } \frac{1}{s_{o}}+\frac{1}{s_{i}}=-\frac{2}{R} \quad \text { or } \quad \frac{1}{30}+\frac{1}{s_{i}}=-\frac{2}{40}\\\ \text { or } s_{i}=-12 \mathrm{~cm} \end{array}$$ The image is virtual \(\left(s_{i}\right.\) is negative) and \(12 \mathrm{~cm}\) behind the mirror. Also, $$M_{T}=-\frac{s_{i}}{s_{o}}=-\frac{-12 \mathrm{~cm}}{30 \mathrm{~cm}}=0.40$$ Moreover, \(M_{T}=y_{i} / y_{o}\) and so \(y_{i}=M_{T} y_{o}=(0.40)(6.0 \mathrm{~m})=2.4 \mathrm{~cm}\)

Describe the image of a candle flame located \(40 \mathrm{~cm}\) from a concave spherical mirror of radius \(64 \mathrm{~cm}\).

What is the focal length of a convex spherical mirror which produces an image one-sixth the size of an object located \(12 \mathrm{~cm}\) from the mirror?

A spherical concave mirror has a radius of curvature of \(-400 \mathrm{~cm}\). An object \(2.00 \mathrm{~cm}\) tall is on the central axis \(400 \mathrm{~cm}\) in front of the mirror. (a) Determine the focal length. (b) Locate the image. ( \(c\) ) Describe the image. (d) Determine the magnification. [Hint: Check out \(\underline{\text { Fig. }} 36-5 .\).]
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