/*! 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 40 You observe an object moving in ... [FREE SOLUTION] | 91Ó°ÊÓ

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You observe an object moving in a straight line and note that at precisely noon the instantaneous velocity is \(45 \mathrm{ft} / \mathrm{sec} .\) Approximately how far has the object moved in the next 5 seconds? Explain carefully what facts learned in this section you are using to obtain your answer.

Short Answer

Expert verified
The object has moved approximately 225 feet in 5 seconds.

Step by step solution

01

Understand the Problem

We are asked to find the approximate distance an object has moved over a period of 5 seconds given its instantaneous velocity at noon is 45 ft/sec. We assume the velocity is constant during this period since no other information is provided.
02

Recognize the Constant Velocity Assumption

Since no information about changing velocity is provided, and the problem involves a short time interval, we assume the velocity remains constant at 45 ft/sec for simplicity.
03

Apply the Constant Velocity Formula

The distance traveled by an object moving at constant velocity is given by the formula: \[\text{Distance} = \text{Velocity} \times \text{Time}\].
04

Calculate the Distance

Using the formula from Step 3 with the given velocity (45 ft/sec) and time period (5 seconds), we calculate the distance:\[\text{Distance} = 45 \, \text{ft/sec} \times 5 \, \text{sec} = 225 \, \text{ft}\].

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Key Concepts

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

Understanding Constant Velocity
When dealing with constant velocity, you can imagine an object moving without speeding up or slowing down. It's like driving a car on a highway at a steady speed without pressing the gas or brake pedals. In physics, constant velocity means that both the speed and direction of an object remain unchanged over time.

This concept simplifies calculations because it allows us to use straightforward formulas. When given an instantaneous velocity (like 45 ft/sec at noon in this exercise), and no information on changes to that speed, it's reasonable to assume the speed remains constant. This assumption is crucial for simplifying our task, as it means the object is moving at the same pace throughout the observed time interval.
Basics of Distance Calculation
To calculate how far an object travels when it moves at a constant velocity, we use the formula:
  • \[ \text{Distance} = \text{Velocity} \times \text{Time} \]
This formula is simple yet powerful. It states that the distance an object travels is equal to how fast it is going (velocity) times how long it travels (time).
In our example, with a constant velocity of 45 ft/sec and a time period of 5 seconds, the calculation becomes:
  • \[ 45 \, \text{ft/sec} \times 5 \, \text{sec} = 225 \, \text{ft} \]
This means the object covers 225 feet in those 5 seconds. This straightforward approach works because the velocity doesn’t change.
Motion in a Straight Line
Motion in a straight line refers to an object that travels in one direction along a single path without turning. This is one of the simplest types of motion and is a common first step in studying movement in physics.
In this exercise, the object moves along a straight line. This simplifies our calculations because we don't have to consider changes in direction or something like circular motion. With only one direction involved, such as forward or backward, equations become much easier to manage.
  • With motion in a straight line, velocity, and distance calculations do not need to account for changes in direction.
By focusing on motion in a straight line, we can learn the basic principles of physics that apply in more complex scenarios too. This kind of motion provides a straightforward context, which helps in understanding how different factors like velocity and time interact to determine distance traveled.

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Most popular questions from this chapter

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Find, without graphing, where each of the given functions is continuous. $$ f(x)=\left\\{\begin{array}{ll} -x+2 & \text { if } x<1 \\ 0 & \text { if } x=1 \\ x^{2} & \text { if } x>1 \end{array}\right. $$

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