/*! 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 16 A mouse travels along a straight... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 16

A mouse travels along a straight line; its distance \(x\) from the origin at any time \(t\) is given by the equation \(x=\left(8.5 \mathrm{cm} \cdot \mathrm{s}^{-1}\right) t-\left(2.5 \mathrm{cm} \cdot \mathrm{s}^{-2}\right) t^{2}\) . Find the average velocity of the mouse in the interval from \(t=0\) to \(t=1.0 \mathrm{s}\) and in the interval from \(t=0\) to \(t=4.0 \mathrm{s}\)

Short Answer

Expert verified
Average velocity from \(t=0\) to \(t=1\) is 6.0 cm/s, and from \(t=0\) to \(t=4\) is -1.5 cm/s.

Step by step solution

01

Understand the Problem

We are given a function describing the distance \(x\) of a mouse from the origin as a function of time \(t\), given by: \(x = (8.5 \text{ cm/s})t - (2.5 \text{ cm/s}^2)t^2\). We need to find the average velocity of the mouse in two different time intervals: from \(t=0\) to \(t=1.0 \text{ s}\) and from \(t=0\) to \(t=4.0 \text{ s}\).
02

Calculate Position at Endpoints

Compute the mouse's position at the beginning and end of each time interval. For \(t=0\): \[x(0) = (8.5)(0) - (2.5)(0)^2 = 0 \text{ cm}\] For \(t=1.0 \text{ s}\): \[x(1) = (8.5)(1) - (2.5)(1)^2 = 8.5 - 2.5 = 6.0 \text{ cm}\] For \(t=4.0 \text{ s}\): \[x(4) = (8.5)(4) - (2.5)(4)^2 = 34 - 40 = -6.0 \text{ cm}\]
03

Calculate Displacement

Determine the displacement for each interval by subtracting the initial position from the final position for each interval. For \([0, 1.0]\): \(\Delta x = x(1) - x(0) = 6.0 - 0 = 6.0 \text{ cm}\). For \([0, 4.0]\): \(\Delta x = x(4) - x(0) = -6.0 - 0 = -6.0 \text{ cm}\).
04

Calculate Time Intervals

Compute the length of each time interval. For \([0, 1.0]\): \(\Delta t = 1.0 - 0 = 1.0 \text{ s}\). For \([0, 4.0]\): \(\Delta t = 4.0 - 0 = 4.0 \text{ s}\).
05

Calculate Average Velocity

Divide the displacement by the time interval to find the average velocity for each interval. Average velocity for \([0, 1.0]\): \[v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{6.0 \text{ cm}}{1.0 \text{ s}} = 6.0 \text{ cm/s}\]. For \([0, 4.0]\): \[v_{\text{avg}} = \frac{-6.0 \text{ cm}}{4.0 \text{ s}} = -1.5 \text{ cm/s}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Displacement
Displacement is the change in position of an object. It tells us the direct distance between the initial and final points, regardless of the path taken. Displacement can be positive, negative, or zero depending on the direction of the movement.
In the given exercise, displacement is calculated for specific time intervals using the position function of the mouse. For the interval from \(t = 0\) to \(t = 1.0 \, \text{s}\), the displacement is \( \Delta x = x(1) - x(0) \), leading to \(\Delta x = 6.0 \, \text{cm}\).
In the interval from \(t = 0\) to \(t = 4.0 \, \text{s}\), the displacement is \(\Delta x = -6.0 \, \text{cm}\).
  • Positive displacement indicates movement away from the starting point.
  • Negative displacement suggests a return or movement toward the starting point.
  • A zero displacement means the final position is the same as the initial position.
Time Interval
A time interval represents the duration between two points in time. Calculating the time interval is essential because it helps us understand how long an event takes.
In the context of average velocity calculations, knowing precisely the time over which the displacement occurred is key. In this problem, the two intervals are:
  • From \(t = 0\) to \(t = 1.0 \, \text{s}\), where \(\Delta t = 1.0 \, \text{s}\)
  • From \(t = 0\) to \(t = 4.0 \, \text{s}\), where \(\Delta t = 4.0 \, \text{s}\)
Each interval shows how much time goes by, providing a necessary component for calculating average velocity.
Velocity Calculation
Velocity calculation involves determining the rate of change in position with respect to time. It is represented as the displacement divided by the time interval and reflects the object's speed and direction over that period.
In the exercise, the average velocity (\(v_{\text{avg}}\)) is calculated as follows:
  • For \([0, 1.0] \, \text{s}\): \(v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{6.0 \, \text{cm}}{1.0 \, \text{s}} = 6.0 \, \text{cm/s}\)
  • For \([0, 4.0] \, \text{s}\): \(v_{\text{avg}} = \frac{-6.0 \, \text{cm}}{4.0 \, \text{s}} = -1.5 \, \text{cm/s}\)
This calculation reveals the consistent rate of motion across time, factoring in both speed and heading.
The positive average velocity in the first interval implies the mouse moved away from the origin, while the negative value in the second suggests a movement in the opposite direction.
Position Function
The position function describes how an object鈥檚 location changes over time. It is the foundation for understanding how displacement and velocity evolve.
In our exercise, the position function of the mouse is given by: \[x(t) = (8.5 \, \text{cm/s})t - (2.5 \, \text{cm/s}^2)t^2\]
This equation considers two parts:
  • \((8.5 \, \text{cm/s})t\) which implies a linear component where speed is constant at \(8.5 \, \text{cm/s}\)
  • \(-(2.5 \, \text{cm/s}^2)t^2\) which suggests the presence of acceleration (negative), decelerating the mouse.
The position function allows us to compute the precise location of the mouse at any given time \(t\). It provides insight into how quickly and in what manner the mouse covers distance from the origin. Using this function, we frame the context for calculating displacement and subsequently average velocity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A helicopter 8.50 m above the ground and descending at 3.50 m/s drops a package from rest (relative to the helicopter). Just as it hits the ground, find (a) the velocity of the package relative to the helicopter and (b) the velocity of the helicopter relative to the package. The package falls freely.

An airplane travels 280 \(\mathrm{m}\) down the runway before taking off. Assuming that it has constant acceleration, if it starts from rest and becomes airborne in 8.00 s, how fast \((\) in \(\mathrm{m} / \mathrm{s})\) is it moving at takeoff?

A fast pitch. The fastest measured pitched baseball left the pitcher's hand at a speed of 45.0 \(\mathrm{m} / \mathrm{s}\) . If the pitcher was in contact with the ball over a distance of 1.50 \(\mathrm{m}\) and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no appreciable air drag. (a) How tall, in meters, is the building? (b) How fast is the brick moving just before it reaches the ground? (c) Sketch graphs of this falling brick's acceleration, velocity, and vertical position as functions of time.

Life from space? One rather controversial suggestion for the origin of life on the earth is that the seeds of life arrived here from outer space in tiny spores-a theory called panspermia. Of course, any life in such spores would have to survive a long journey through the frigid near vacuum of outer space. If these spores originated at the distance of our nearest star, Alpha Centauri, and traveled at a constant \(1000 \mathrm{km} / \mathrm{s},\) how many years would it take for them to make the journey here? (Alpha Centauri is approximately 4\(\frac{1}{4}\) light years away, and the light year is the distance light travels in 1 year.)
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Fields in Physics

Read Explanation

Dynamics

Read Explanation

Kinematics Physics

Read Explanation

Work Energy and Power

Read Explanation

Electromagnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.