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Chapter 2: Problem 1

What is the magnitude of the displacement of a car that travels half a lap along a circle that has a radius of \(150 \mathrm{~m} ?\) How about when the car travels a full lap?

Short Answer

Expert verified
Half a lap: 300 m; Full lap: 0 m.

Step by step solution

01

Understand the Problem

We need to find the displacement, which is the shortest distance from the initial to the final point. The car travels half a lap and then a full lap along a circle with a given radius.
02

Identify Key Formulas

For a circle, the diameter is twice the radius, and the displacement after half a lap is the diameter. The circumference of the circle is used for the displacement of a full lap. The formula for circumference is \( C = 2\pi r \).
03

Calculate the Diameter

The diameter of the circle is twice the radius. Given that the radius \( r = 150 \text{ m} \), the diameter \( d = 2r = 2 \times 150 = 300 \text{ m} \). This is the displacement for half a lap.
04

Calculate the Full Lap Displacement

When the car completes a full lap, it returns to the starting point. Therefore, the displacement is 0 m because displacement depends only on the initial and final positions.

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Key Concepts

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

Radius of a Circle
Understanding the concept of the radius is crucial in grasping circular motion. The radius of a circle is defined as the distance from the center of the circle to any point on its boundary. It helps determine other fundamental properties of the circle.

The radius is typically denoted by the letter \( r \) and is integral to calculations involving circles, such as diameter and circumference. Knowing the radius allows you to visualize the circle's size and explore different aspects of circular movement.
  • **Key Feature**: Every point on a circle is equidistant from the center.
  • **Role in Calculations**: The radius is often used to calculate the diameter (\(d = 2r\)) and the circumference (\(C = 2\pi r\)).
Whenever dealing with problems involving circles, it's helpful to first determine the radius to guide further analysis.
Circumference of a Circle
The circumference is the distance around the boundary of a circle. It can be thought of as the circle's perimeter. When calculating the circumference, remember that it represents the linear distance completed when tracing around the circle once.

The standard formula for the circumference of a circle is:\[C = 2\pi r\]Where \( C \) is the circumference and \( r \) is the radius. This formula highlights the relationship between a circle's radius and its circumference. The value of \( \pi \) is approximately 3.14159, and it emerges naturally when working with circles.
  • **Significance in Motion**: Knowing the circumference is essential for understanding how far an object travels along the circle's path.
  • **Calculation**: Simply double the radius and multiply by \( \pi \) to find the circumference.
The circumference plays a fundamental role in any problem involving circular paths, providing the basis for understanding travel distance when moving around the circle.
Diameter of a Circle
The diameter of a circle is a straight line passing through the center and touching two points on its boundary. It effectively divides the circle into two equal halves. This property makes the diameter twice the length of the radius, expressed symbolically as:
\[d = 2r\]
In practical terms, understanding the diameter is essential when solving geometric problems involving circles.
  • **Geometry**: The diameter is not only twice the radius but also the longest chord in a circle.
  • **Application in Displacement**: For circular motion, the diameter can determine the straight-line distance (displacement) after covering half of the circle.
In exercises like the one provided, knowing how to calculate and apply the diameter is vital for accurately determining displacement.

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Most popular questions from this chapter

A couple is traveling by car down a straight highway at \(40 \mathrm{~km} / \mathrm{h}\). They see an accident in the distance, so the driver applies the brakes, and in \(5.0 \mathrm{~s}\) the car uniformly slows down to rest. (a) The direction of the acceleration vector is (1) in the same direction as, (2) opposite to, (3) at \(90^{\circ}\) relative to the velocity vector. Why? (b) By how much must the velocity change each second from the start of braking to the car's complete stop?

Two identical cars capable of accelerating at \(3.00 \mathrm{~m} / \mathrm{s}^{2}\) are racing on a straight track with running starts. Car \(\mathrm{A}\) has an initial speed of \(2.50 \mathrm{~m} / \mathrm{s} ;\) car \(\mathrm{B}\) starts with speed of \(5.00 \mathrm{~m} / \mathrm{s}\). (a) What is the separation of the two cars after \(10 \mathrm{~s} ?\) (b) Which car is moving faster after \(10 \mathrm{~s} ?\)

The displacement of an object is given as a function of time by \(x=3 t^{2} \mathrm{~m} .\) What is the magnitude of the average velocity for \((\) a) \(\Delta t=2.0 \mathrm{~s}-0 \mathrm{~s},\) and (b) \(\Delta t=4.0 \mathrm{~s}-2.0 \mathrm{~s} ?\)

A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is \(25.0 \mathrm{~m}\) behind the car (v Fig. 2.27). The car accelerates at a uniform rate of \(3.70 \mathrm{~m} / \mathrm{s}^{2}\) and the motorcycle at a uniform rate of \(4.40 \mathrm{~m} / \mathrm{s}^{2}\) (a) How much time elapses before the motorcycle overtakes the car? (b) How far will each have traveled during that time? (c) How far ahead of the car will the motorcycle be \(2.00 \mathrm{~s}\) later? (Both vehicles are still accelerating.)

In the 1960 s there was a contest to find the car that could do the following two maneuvers (one right after the other) in the shortest total time: First, accelerate from rest to \(100 \mathrm{mi} / \mathrm{h}(45.0 \mathrm{~m} / \mathrm{s}),\) and then brake to a complete stop. (Ignore the reaction time correction that occurs between the speeding-up and slowing-down phases and assume that all accelerations are constant.) For several years, the winner was the "James Bond car," the Aston Martin. One year it won the contest when it took a total of only 15.0 seconds to perform these two tasks! Its braking acceleration (deceleration) was known to be an excellent \(9.00 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the time it took during the braking phase. (b) Calculate the distance it traveled during the braking phase. (c) Calculate the car's acceleration during the speeding-up phase. (d) Calculate the distance it took to reach \(100 \mathrm{mi} / \mathrm{h}\).
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