/*! 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 43 a. What constant acceleration, i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 43

a. What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? b. What fraction of \(g\) is this? c. How far has the car traveled when it reaches 60 mph? Give your answer both in SI units and in feet.

Short Answer

Expert verified
The car must have a constant acceleration of approximately \(2.68224 m/s^2\), which is about \(0.274\) of \(g\). The car would have traveled about \(134.112 m\) or \(439.963 ft\) when it reaches \(60 mph\).

Step by step solution

01

Convert Speed from mph to m/s

Make sure all units are in SI units for ease of calculation. The given final speed of the car is 60 mph. Convert this to metres per second (m/s) using the conversion factor \(1 mph = 0.44704 m/s\). Therefore,\( 60 mph = 60 * 0.44704 m/s = 26.8224 m/s\).
02

Apply First Equation of Motion

The first equation of motion is \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is acceleration and \(t\) is time. Here, \(u = 0 m/s\) because the car starts from rest, \(v = 26.8224 m/s\) and \(t = 10s\). Solve the equation for \(a\): \(a = (v-u) / t = 26.8224 m/s 梅 10s = 2.68224 m/s^2\).
03

Express Acceleration as a Fraction of Gravitational Constant

The acceleration due to gravity \(g\) is approximately \(9.8 m/s^2\). Compute \(a/g\) to get the fraction of acceleration to gravity: \(a/g = 2.68224 m/s^2 梅 9.8 m/s^2 \approx 0.274\). Thus, the acceleration is approximately 0.274 of the gravitational constant.
04

Calculate the Displacement Using Second Equation of Motion

Using the second equation of motion \(s = ut + 0.5at^2\), where \(s\) is the displacement, compute for \(s\): \(s = ut + 0.5at^2 = 0 (as u = 0) + 0.5 x 2.68224 m/s^2 x (10 s)^2 = 134.112 m\).
05

Convert Displacement from Meters to Feet

Use the conversion factor \(1 m = 3.281 ft\) to convert \(134.112 m\) to feet: \(134.112 m * 3.281 ft/m = 439.963 ft\).

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.

Units Conversion
When solving physics problems, one must often convert units to ensure consistency in equations. For velocity, a common conversion is from miles per hour (mph) to meters per second (m/s), because m/s is the standard in the International System of Units (SI). To do this, use the conversion factor:
  • 1 mph = 0.44704 m/s
For example, to convert 60 mph to m/s:
  • Calculate: 60 mph 脳 0.44704 m/s = 26.8224 m/s
This ensures that calculations involving speed are aligned with SI unit measurements, such as time in seconds or distance in meters. Converting units correctly is crucial for the success of solving physics problems.
Equations of Motion
The equations of motion allow us to calculate various components of an object's movement, assuming constant acceleration. They help to determine quantities like velocity, displacement, and time. The first equation of motion is:
  • \( v = u + at \)
Where:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is acceleration
  • \( t \) is time
In our problem, the car starts from rest, so \( u = 0 \). If \( v = 26.8224 \) m/s and \( t = 10 \) s, we can solve for \( a \). The second equation of motion is:
  • \( s = ut + 0.5at^2 \)
Here, \( s \) represents the displacement, allowing us to calculate how far an object has traveled over time. Apply these equations step by step for calculating various aspects of motion under a constant acceleration.
Gravitational Force
Gravitational force is the attraction between two masses. On Earth, this force results in the acceleration known as gravity, denoted as \( g \), which is approximately 9.8 m/s虏. When comparing other accelerations to gravity, one may express acceleration as a fraction of \( g \). This offers perspective on how the force or acceleration compares to what we experience due to gravity daily.For instance, if an object has an acceleration of 2.68224 m/s虏, it can be compared to \( g \) by:
  • Calculating: \( a/g = 2.68224 \ m/s^2 / 9.8 \ m/s^2 \approx 0.274 \)
This means the object's acceleration is 0.274 times that of gravitational acceleration.
Distance Displacement
Distance and displacement describe an object's movement but are different concepts. Distance refers to the entire path traveled, while displacement measures the shortest distance from the start to end point. Using the equations of motion to find displacement involves:
  • Starting with: \( s = ut + 0.5at^2 \)
For a car starting from rest, and with a calculated acceleration, \( a \), over a time, \( t \), you find \( s \), the displacement. It's important for physics calculations to know exactly where and how far an object has moved, particularly in studies involving forces and motion.
Metric Conversion
Metric conversion allows us to switch between different units of measurement within the metric system or convert between metric and imperial units, such as meters to feet. Understanding these conversions is essential for accurately interpreting and converting physical quantities. For example, converting a displacement from meters to feet uses:
  • The conversion factor: 1 m = 3.281 ft
Using this:
  • 134.112 meters becomes: 134.112 m 脳 3.281 ft/m = 439.963 ft
Making this conversion helps to contextualize results in a unit that might be more familiar to those accustomed to imperial measurements.

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

I was driving along at \(20 \mathrm{m} / \mathrm{s}\), trying to change a CD and not watching where I was going. When I looked up, I found myself \(45 \mathrm{m}\) from a railroad crossing. And wouldn't you know it, a train moving at \(30 \mathrm{m} / \mathrm{s}\) was only \(60 \mathrm{m}\) from the crossing. In a split second, I realized that the train was going to beat me to the crossing and that I didn't have enough distance to stop. My only hope was to accelerate enough to cross the tracks before the train arrived. If my reaction time before starting to accelerate was \(0.50 \mathrm{s},\) what minimum acceleration did my car need for me to be here today writing these words?

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 s later. What was the rocket's acceleration?

You are given the kinematic equation or equations that are used to solve a problem. For each of these, you are to: a. Write a realistic problem for which this is the correct equation(s). Be sure that the answer your problem requests is consistent with the equation(s) given. b. Draw the pictorial representation for your problem. c. Finish the solution of the problem. $$64 \mathrm{m}=0 \mathrm{m}+(32 \mathrm{m} / \mathrm{s})(4 \mathrm{s}-0 \mathrm{s})+\frac{1}{2} a_{x}(4 \mathrm{s}-0 \mathrm{s})^{2}$$

A car starts from rest at a stop sign. It accelerates at \(4.0 \mathrm{m} / \mathrm{s}^{2}\) for \(6.0 \mathrm{s},\) coasts for \(2.0 \mathrm{s},\) and then slows down at a rate of \(3.0 \mathrm{m} / \mathrm{s}^{2}\) for the next stop sign. How far apart are the stop signs?

You are driving to the grocery store at \(20 \mathrm{m} / \mathrm{s}\). You are \(110 \mathrm{m}\) from an intersection when the traffic light turns red. Assume that your reaction time is \(0.50 \mathrm{s}\) and that your car brakes with constant acceleration. a. How far are you from the intersection when you begin to apply the brakes? b. What acceleration will bring you to rest right at the intersection? c. How long does it take you to stop after the light turns red?
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Solid State Physics

Read Explanation

Wave Optics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Turning Points in Physics

Read Explanation

Waves 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.