/*! 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 74 A runner of mass \(56.1 \mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 74

A runner of mass \(56.1 \mathrm{~kg}\) starts from rest and accelerates with a constant acceleration of \(1.23 \mathrm{~m} / \mathrm{s}^{2}\) until she reaches a velocity of \(5.10 \mathrm{~m} / \mathrm{s}\). She then continues running at this constant velocity. How long does the runner take to travel \(173 \mathrm{~m} ?\)

Short Answer

Expert verified
Answer: To find the total time taken by the runner to travel 173 meters, follow these steps: 1. Calculate the time for the acceleration phase using \(t = \frac{v - u}{a}\). 2. Calculate the distance traveled during acceleration using \(s = ut + \frac{1}{2} at^2\). 3. Calculate the remaining distance to be covered by subtracting the distance traveled during acceleration from 173 meters. 4. Calculate the time for the constant velocity phase using \(t = \frac{s}{v}\). 5. Add the time spent during the acceleration and constant velocity phases to find the total time required for the runner to travel 173 meters.

Step by step solution

01

Calculate the time for the acceleration phase

First, we need to find how long the runner takes to reach the constant velocity of \(5.10 \mathrm{~m} / \mathrm{s}\). We can use the equation \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity (which is \(0\) since the runner starts from rest), \(a\) is the acceleration, and \(t\) is the time: \begin{equation} t = \frac{v - u}{a} \end{equation} Plugging in values: \(v = 5.10 \mathrm{~m/s}, u = 0 \mathrm{~m/s}\) and \(a = 1.23 \mathrm{~m/s^2}\), we'll get the time for the acceleration phase.
02

Calculate the distance traveled during acceleration

Next, we need to find the distance traveled during the acceleration phase. We'll use the equation \(s = ut + \frac{1}{2} at^2\), where \(s\) is the distance traveled, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time found in Step 1. Plugging in values: \(u = 0 \mathrm{~m/s}, a = 1.23 \mathrm{~m/s^2}\), and \(t\) from Step 1, we'll find the distance traveled during acceleration.
03

Calculate the distance remaining to be covered

Now, we need to find the remaining distance the runner needs to cover before reaching \(173 \mathrm{~m}\). We can do this by subtracting the distance traveled during acceleration (found in Step 2) from \(173 \mathrm{~m}\).
04

Calculate the time for the constant velocity phase

Now, we'll find the time required to cover the remaining distance at a constant velocity of \(5.10 \mathrm{~m/s}\). We can use the equation \(t = \frac{s}{v}\), where \(t\) is the time, \(s\) is the distance remaining (found in Step 3), and \(v\) is the constant velocity.
05

Calculate the total time the runner takes to travel \(173 \mathrm{~m}\)

Finally, we'll add the time spent during the acceleration phase (Step 1) and the time spent during the constant velocity phase (Step 4) to find the total time required for the runner to travel \(173 \mathrm{~m}\).

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.

Constant Acceleration
Understanding constant acceleration is crucial in solving problems related to motion. An object experiences constant acceleration when its velocity changes at a uniform rate over time. This means the object's speed increases by the same amount every second. For example, in the exercise, the runner starts from rest and changes her velocity uniformly due to constant acceleration.

This concept is described by the equation:
  • \( v = u + at \)
Where:
  • \( v \) is the final velocity.
  • \( u \) is the initial velocity (which is 0 if starting from rest).
  • \( a \) is the acceleration.
  • \( t \) is the time taken to reach the final velocity.
This formula helps us determine how long it will take the runner to reach a certain speed given a specific acceleration. By using this formula, we calculated how long the runner takes to accelerate from a standstill to 5.10 m/s.
Velocity
Velocity, a key concept in kinematics, is the speed of an object in a particular direction. It's important to distinguish velocity from speed because velocity includes direction, making it a vector quantity. In our exercise, once the runner reaches a velocity of 5.10 m/s, she maintains this speed as she continues running.

Velocity is determined by the equation:
  • \( v = \frac{s}{t} \)
Where:
  • \( v \) is the velocity.
  • \( s \) is the distance traveled.
  • \( t \) is the time taken to travel that distance.
Understanding velocity allows us to solve how long it takes to cover a particular distance while maintaining a constant speed. This is used in the exercise to find the duration of time the runner needs to maintain her velocity to cover the remaining distance after accelerating.
Distance-Time Relationship
The relationship between distance and time is central to solving motion problems. It tells us how the distance an object travels changes over time, depending on whether it's accelerating or moving at a constant speed.

In the context of our exercise, two main scenarios involve this relationship:
  • During the acceleration phase, the distance is given by the formula:
    • \( s = ut + \frac{1}{2} at^2 \)
  • During the constant velocity phase, the distance is given by the formula:
    • \( s = vt \)
In these formulas:
  • \( s \) is the distance traveled.
  • For the acceleration phase, initial velocity \( u \) is zero, \( a \) is the acceleration, and \( t \) is the time to reach final velocity.
  • For the constant motion phase, \( v \) is the constant velocity, and \( t \) is the time to cover the remaining distance.
By using these formulas, we can accurately calculate the total distance the runner travels during each phase of her run and thus determine the total time to cover 173 meters.

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

You drop a water balloon straight down from your dormitory window \(80.0 \mathrm{~m}\) above your friend's head. At \(2.00 \mathrm{~s}\) after you drop the balloon, not realizing it has water in it your friend fires a dart from a gun, which is at the same height as his head, directly upward toward the balloon with an initial velocity of \(20.0 \mathrm{~m} / \mathrm{s}\). a) How long after you drop the balloon will the dart burst the balloon? b) How long after the dart hits the balloon will your friend have to move out of the way of the falling water? Assume the balloon breaks instantaneously at the touch of the dart.

The position of a particle moving along the \(x\) -axis is given by \(x=\left(11+14 t-2.0 t^{2}\right),\) where \(t\) is in seconds and \(x\) is in meters. What is the average velocity during the time interval from \(t=1.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s} ?\)

A jet touches down on a runway with a speed of \(142.4 \mathrm{mph} .\) After \(12.4 \mathrm{~s},\) the jet comes to a complete stop. Assuming constant acceleration of the jet, how far down the runway from where it touched down does the jet stand?

A ball is thrown straight upward in the air at a speed of \(15.0 \mathrm{~m} / \mathrm{s} .\) Ignore air resistance. a) What is the maximum height the ball will reach? b) What is the speed of the ball when it reaches \(5.00 \mathrm{~m} ?\) c) How long will it take to reach \(5.00 \mathrm{~m}\) above its initial position on the way up? d) How long will it take to reach \(5.00 \mathrm{~m}\) above its initial position on its way down?

A girl is riding her bicycle. When she gets to a corner, she stops to get a drink from her water bottle. At that time, a friend passes by her, traveling at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). a) After \(20 \mathrm{~s}\), the girl gets back on her bike and travels with a constant acceleration of \(2.2 \mathrm{~m} / \mathrm{s}^{2} .\) How long does it take for her to catch up with her friend? b) If the girl had been on her bike and rolling along at a speed of \(1.2 \mathrm{~m} / \mathrm{s}\) when her friend passed, what constant acceleration would she need to catch up with her friend in the same amount of time?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Classical Mechanics

Read Explanation

Electromagnetism

Read Explanation

Electricity

Read Explanation

Medical Physics

Read Explanation

Astrophysics

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.