/*! 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 18 A car's velocity as a function o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 18

A car's velocity as a function of time is given by \(v_{x}(t)=\) \(\alpha+\beta t^{2},\) where \(\alpha=3.00 \mathrm{m} / \mathrm{s}\) and \(\beta=0.100 \mathrm{m} / \mathrm{s}^{3}\) , (a) Calculate the average acceleration for the time interval \(t=0\) to \(t=5.00 \mathrm{s}\) (b) Calculate the instantaneous acceleration for \(t=0\) and \(t=\) 5.00 \(\mathrm{s}\) . (c) Draw accurate \(v_{x}-t\) and \(a_{x}-t\) graphs for the car's motion between \(t=0\) and \(t=5.00 \mathrm{s} .\)

Short Answer

Expert verified
(a) 0.500 m/s虏; (b) 0.00 m/s虏 at t=0, 1.00 m/s虏 at t=5.00 s.

Step by step solution

01

Understanding Velocity Function

The velocity function is given as \( v_{x}(t) = \alpha + \beta t^{2} \) where \( \alpha = 3.00 \, \text{m/s} \) and \( \beta = 0.100 \, \text{m/s}^3 \). This shows that the velocity is a function of time that increases quadratically.
02

Calculate Average Acceleration

The average acceleration \( a_{\text{avg}} \) over a time interval is calculated by the change in velocity divided by the change in time. \[a_{\text{avg}} = \frac{v_{x}(t_{2}) - v_{x}(t_{1})}{t_{2} - t_{1}}\] Substituting \( t_{1} = 0 \) and \( t_{2} = 5.00 \, \text{s} \), calculate the velocities: \( v_{x}(0) = 3.00 \, \text{m/s} \) and \( v_{x}(5.00) = 3.00 + 0.100 \times 5^{2} = 5.50 \, \text{m/s} \). Thus, \[ a_{\text{avg}} = \frac{5.50 - 3.00}{5 - 0} = \frac{2.50}{5} = 0.500 \, \text{m/s}^2 \] The average acceleration is \( 0.500 \, \text{m/s}^2 \).
03

Determine Instantaneous Acceleration

Instantaneous acceleration is the derivative of velocity with respect to time. \[a_{x}(t) = \frac{d}{dt}(v_{x}(t)) = \frac{d}{dt}(\alpha + \beta t^{2}) = 2\beta t \]Substitute \( t = 0 \): \[ a_{x}(0) = 2 \times 0.100 \times 0 = 0.00 \, \text{m/s}^2 \] Substitute \( t = 5.00 \): \[ a_{x}(5.00) = 2 \times 0.100 \times 5 = 1.00 \, \text{m/s}^2 \] The instantaneous accelerations are \( 0.00 \, \text{m/s}^2 \) at \( t = 0 \) and \( 1.00 \, \text{m/s}^2 \) at \( t = 5.00 \text{s} \) respectively.
04

Sketch VT and AT Graphs

For \( v_{x}-t \) graph, plot \( v_{x}(t) = 3.00 + 0.100t^{2} \), showing a parabola opening upwards starting from \( 3.00 \text{m/s} \) at \( t=0 \) to \( 5.50 \text{m/s} \) at \( t=5.00 \text{s} \). For \( a_{x}-t \) graph, plot \( a_{x}(t) = 0.200t \), starting at \( 0.00 \text{m/s}^2 \) at \( t=0 \) to \( 1.00 \text{m/s}^2 \) at \( t=5.00 \text{s} \). This line shows linear increase.

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.

Average Acceleration Calculation
The average acceleration of an object is a measure of how quickly its velocity changes over a specific time interval. To find the average acceleration, use the formula:
  • \(a_{\text{avg}} = \frac{v_{x}(t_{2}) - v_{x}(t_{1})}{t_{2} - t_{1}}\)

Look at the car's velocity function, which is \(v_{x}(t) = 3.00 + 0.100 t^{2}\). Here, you calculate the velocity at two different points in time.
For the interval from \(t=0\) to \(t=5.00\) seconds, the velocity changes from \(3.00 \, \text{m/s}\) to \(5.50 \, \text{m/s}\).
This results in an average acceleration of \(0.500 \, \text{m/s}^2\). Understanding how velocity changes over time is key to grasping average acceleration concepts.
Instantaneous Acceleration
Instantaneous acceleration is the rate at which velocity changes at a specific moment. Unlike average acceleration, it's not an overall rate over a period, but a snapshot at a precise time.
To find the instantaneous acceleration, take the derivative of the velocity function with respect to time. For the velocity function \(v_{x}(t) = 3.00 + 0.100 t^{2}\), differentiate to get:
  • \(a_{x}(t) = 2 \beta t = 0.200t\)

Evaluate at specific times to find:
  • At \(t=0\), \(a_{x}(0) = 0.00 \, \text{m/s}^2\).
  • At \(t=5.00\), \(a_{x}(5.00) = 1.00 \, \text{m/s}^2\).
This means the acceleration grows linearly with time. Grasping the concept of instantaneous acceleration helps you understand how quickly an object's speed is changing right at any given moment.
Velocity-Time Graph
The velocity-time graph is a crucial tool for visualizing how an object's velocity changes over time. For this problem, the equation \(v_{x}(t) = 3.00 + 0.100 t^{2}\) illustrates a parabolic curve. This is because the velocity is a quadratic function of time.
Here's what the graph tells you:
  • At \(t=0\), the car's velocity is \(3.00 \, \text{m/s}\).
  • The velocity increases to \(5.50 \, \text{m/s}\) at \(t=5.00\).

This increasing pattern, shown as a parabola opening upwards, indicates the car is accelerating more rapidly as time progresses.
A velocity-time graph helps in easily interpreting the rate of changes in motion and can be used to derive important details like displacement and change in speed.
Acceleration-Time Graph
An acceleration-time graph displays how an object's acceleration changes over time. For this car scenario, the acceleration function is linear, derived as \(a_{x}(t) = 0.200t\). This equation implies a consistent increase in acceleration as time goes on.
What to note from the graph:
  • At \(t=0\), acceleration is \(0.00 \, \text{m/s}^2\).
  • At \(t=5.00\), acceleration reaches \(1.00 \, \text{m/s}^2\).

On the graph, this is represented by a straight line sloping upwards. This linearity captures a uniform rate of change of velocity with respect to time.
Understanding the acceleration-time graph provides insight into how rapidly the velocity of an object is changing at any given moment, reinforcing the understanding of acceleration dynamics.

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 rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after Iftoff, the rocket clears the top of its launch platform, 63 \(\mathrm{m}\) above the ground. After an additional \(4.75 \mathrm{s},\) it is 1.00 \(\mathrm{km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 -s part of its flight and (b) the first 5.90 s of its flight.

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50.0 \(\mathrm{m}\) below its starting point 5.00 s after it leaves the thrower's hand. Air resistance may be ignored. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the egg.

You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 m/s passes you. When the rear of the bus is 12.0 m past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 m/s. How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

Auto Acceleration. Based on your experiences of riding in antomobiles, estimate the magnitude of a car's average acceleration when it (a) accelerates onto a freeway from rest to 65 \(\mathrm{mi} / \mathrm{h}\) , and (b) brakes from highway speeds to a sudden stop. (c) Explain why the average acceleration in each case could be regarded as either positive or negative.

An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.30 s for the boulder to fall the last third of the way to the ground. You may ignore air resistance. (a) What is the height of the cliff in meters? (b) If in part (a) you get two solutions of a quadratic equation and you use one for your answer, what does the other solution represent?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Classical Mechanics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Magnetism

Read Explanation

Thermodynamics

Read Explanation

Engineering Physics

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.