/*! 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 45 You can run \(9.0 \mathrm{m} / \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 45

You can run \(9.0 \mathrm{m} / \mathrm{s}, 20 \%\) faster than your brother. How much head start should you give him in order to have a tie race over \(100 \mathrm{m} ?\)

Short Answer

Expert verified
The head start that you should give to your brother in order to have a tie race over 100 meters is approximately 2.22 seconds.

Step by step solution

01

Calculate the brother鈥檚 speed

The current speed is \(9.0 \mathrm{m}/\mathrm{s}\) and it is given that it is 20% faster than the brother's speed, so his speed (let's name it \(V_{b}\) to reference it later) will be: \(V_{b} = 9.0 \mathrm{m} / \mathrm{s} / 1.2 = 7.5 \mathrm{m} / \mathrm{s}\)
02

Calculate the time it takes for the brother to finish

Using the formula Time = Distance / Speed, the time (let's name it \(T_{b}\) to reference it later) it would take the brother to finish the 100 m race is calculated as: \(T_{b} = 100 \mathrm{m} / V_{b} = 100 \mathrm{m} / 7.5 \mathrm{m} / \mathrm{s} = 13.33 \mathrm{s}\)
03

Calculate the head start

Now to tie the race, the time taken by both should be the same. Therefore, the head start (Time\_start) should be calculated as: Time\_start = \(T_{b} - 100 \mathrm{m} / 9.0 \mathrm{m} / \mathrm{s} = 13.33 \mathrm{s} - 11.11 \mathrm{s} = 2.22 \mathrm{s}\)

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!

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're at mission control for a rocket launch, deciding whether to let the launch proceed. A band of clouds \(5.3 \mathrm{km}\) thick extends upward from \(1.9 \mathrm{km}\) altitude. The rocket will accelerate at \(4.6 \mathrm{m} / \mathrm{s}^{2}\) and it isn't allowed to be out of sight for more than 30 s. Should you allow the launch?

You're an investigator for the National Transportation Safety Board, examining a subway accident in which a train going at \(80 \mathrm{km} / \mathrm{h}\) collided with a slower train traveling in the same direction at \(25 \mathrm{km} / \mathrm{h}\). Your job is to determine the relative speed of the collision, to help establish new crash standards. The faster train's "black box" shows that its brakes were applied and it began slowing at the rate of \(2.1 \mathrm{m} / \mathrm{s}^{2}\) when it was \(50 \mathrm{m}\) from the slower train, while the slower train continued at constant speed. What do you report?

Starting from rest, a car accelerates at a constant rate, reaching \(88 \mathrm{km} / \mathrm{h}\) in \(12 \mathrm{s} .\) Find (a) its acceleration and (b) how far it goes in this time.

What's the conversion factor from meters per second to miles per hour?

A model rocket is launched straight upward. Its altitude \(y\) as a function of time is given by \(y=b t-c t^{2},\) where \(b=82 \mathrm{m} / \mathrm{s}, c=4.9 \mathrm{m} / \mathrm{s}^{2}, t\) is the time in seconds, and \(y\) is in meters. (a) Use differentiation to find a general expression for the rocket's velocity as a function of time. (b) When is the velocity zero?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Electricity

Read Explanation

Thermodynamics

Read Explanation

Classical Mechanics

Read Explanation

Famous Physicists

Read Explanation

Conservation of Energy and Momentum

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.