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You are in your car driving on a highway at 25 m\(/\)s when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 cm) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 m\(/\)s when the truck is 2.0 m from the mirror, what is the speed of the truck relative to the highway?

Short Answer

Expert verified
The truck's speed relative to the highway is 50.58 m/s.

Step by step solution

01

Understanding the Problem

We need to calculate the speed of a truck that's moving towards a convex mirror on a car traveling at 25 m/s. The mirror has a radius of curvature of 150 cm, and the image speed is 1.9 m/s when the truck is 2.0 m away from the mirror.
02

Convert Units Wherever Necessary

The given radius of curvature is 150 cm which needs to be converted into meters. So, \( R = 150 \text{ cm} = 1.5 \text{ m} \). This conversion is essential for calculating the focal length and using it in lens/mirror equations.
03

Calculate the Focal Length of the Convex Mirror

For a convex mirror, the focal length \( f \) is given by \( f = \frac{R}{2} \). Hence, \( f = \frac{1.5}{2} = 0.75 \text{ m} \).
04

Use the Mirror Formula

The mirror formula is given by \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \( v \) is the image distance, and \( u \) is the object distance. The truck is 2.0 m from the mirror, so \( u = -2.0 \text{ m} \) (negative for mirrors).
05

Find Image Distance \( v \)

Substitute the known values in the mirror formula: \[ \frac{1}{0.75} = \frac{1}{v} - \frac{1}{2} \]. Solving for \( \frac{1}{v} \) gives \( \frac{1}{v} = \frac{1}{0.75} + \frac{1}{2} = \frac{4}{3} + \frac{1}{2} = \frac{11}{6} \), so \( v = \frac{6}{11} \approx 0.545 \text{ m} \).
06

Use Image Speed Formula

The relation between velocities for mirrors is given by \( \frac{v_i}{v_o} = \left( \frac{v^2}{u^2} \right) \), where \( v_i \) is the image speed (1.9 m/s) and \( v_o \) is the object speed needed. Given \( u = -2 \text{ m} \) and \( v \approx 0.545 \text{ m} \), the ratio \( \left( \frac{0.545^2}{2^2} \right) = \left( \frac{0.297}{4} \right) = 0.07425 \).
07

Solve for Truck's Speed

Using the image speed formula, \( v_o = \frac{v_i}{0.07425} = \frac{1.9}{0.07425} \approx 25.58 \text{ m/s} \). Since the car is moving at 25 m/s, the speed of the truck relative to the highway is \( 25.58 + 25 = 50.58 \text{ m/s} \).

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Key Concepts

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

Radius of Curvature
The radius of curvature is a fundamental concept in understanding mirrors. It refers to the distance from the mirror's surface to the center of the sphere it is part of. For convex mirrors, which curve outward, this radius helps us determine other properties like focal length.
In our exercise, the radius of curvature given is 150 cm. To align with standard physics unit practices, we must convert it to meters by dividing by 100, resulting in 1.5 m. This conversion is critical, as it ensures our calculations for focal length and other attributes are consistent and accurate.
Mirror Formula
The mirror formula is a central equation in optics used to relate the object distance, image distance, and focal length of spherical mirrors. It’s mathematically expressed as: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Here, \( f \) stands for the focal length, \( v \) for image distance, and \( u \) for object distance. A trick to remember this is to think of \( u \) as the distance where the object "U" resides, and \( v \) where the "View" (image) forms.
In solving mirror problems, unless specified otherwise, object distances are considered negative for real objects placed in front of the mirror. In our trailer example, it's crucial to implement these sign conventions accurately to solve the problem.
Focal Length
In the context of spherical mirrors like the convex mirror in the exercise, the focal length \( f \) is half the radius of curvature. Mathematically, it is described as: \[ f = \frac{R}{2} \] For a convex mirror, the focal point is virtual and located behind the mirror, leading to a positive focal length value.
Applying this concept, with a radius of curvature \( R \) of 1.5 m, the focal length becomes \( f = \frac{1.5}{2} = 0.75 \text{ m} \). Understanding focal length allows one to determine how strongly the mirror converges or diverges light and forms images.
Relative Speed Calculation
Analyzing the relative speed between two moving objects requires understanding motion and velocity principles. When considering this truck approaching our mirror in a moving vehicle, relative speed tells us how fast one object approaches or recedes from another.
In the original problem, we calculated the truck’s speed specifically regarding the mirror’s image. First, we determined the image velocity along the mirror formula path. Then, to find the actual truck's speed on the highway relative to a stationary observer, we added the car’s highway speed to the image's relative speed derived from mirror reflections, obtaining \( 50.58 \text{ m/s} \). This approach highlights the need to consider both the image's motion and the overall system motion in calculating relative speed effectively.

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Most popular questions from this chapter

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_1\) = +12.0 cm and \(R_2\) = +28.0 cm. The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first.

Where must you place an object in front of a concave mirror with radius \(R\) so that the image is erect and \(2\over12\) times the size of the object? Where is the image?

A transparent liquid fills a cylindrical tank to a depth of 3.60 m. There is air above the liquid. You look at normal incidence at a small pebble at the bottom of the tank. The apparent depth of the pebble below the liquid's surface is 2.45 m. What is the refractive index of this liquid?

The cornea of the eye has a radius of curvature of approximately 0.50 cm, and the aqueous humor behind it has an index of refraction of 1.35. The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 mm. (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were 25 cm in front of the eye? If not, where would it focus that text: in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about 5.0 mm, where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

A screen is placed a distance \(d\) to the right of an object. A converging lens with focal length \(f\) is placed between the object and the screen. In terms of \(f\), what is the smallest value \(d\) can have for an image to be in focus on the screen?

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