Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 75

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?

Short Answer

Expert verified
Answer: The jet stands about 787.44 meters down the runway after decelerating to a complete stop.

Step by step solution

01

Identify the known values

The problem provides the following information: - Initial velocity of the jet, \(v_i = 142.4\,\mathrm{mph}\) - Time of acceleration, \(t = 12.4\,\mathrm{s}\) - Final velocity, \(v_f = 0\) since the jet comes to a complete stop.
02

Convert the initial velocity to \(\mathrm{m/s}\)

Since we have the time in seconds, we need to convert the initial velocity from \(\mathrm{mph}\) to \(\mathrm{m/s}\): $$v_i = 142.4\,\mathrm{mph} \cdot 1609.34 \mathrm{m/mile} \cdot \dfrac{1\,\mathrm{h}}{3600\,\mathrm{s}} ≈ 63.6\,\mathrm{m/s}$$
03

Find the acceleration

Using the formula \(v_f = v_i + at\), we can solve for acceleration \(a\): $$a = \dfrac{v_f - v_i}{t} = \dfrac{0 - 63.6\,\mathrm{m/s}}{12.4\,\mathrm{s}} ≈ -5.14\,\mathrm{m/s^2}$$
04

Find the distance traveled

Finally, use the equation \(d = v_i t + 0.5at^2\) to find the distance the jet traveled on the runway: $$d = 63.6\,\mathrm{m/s}\cdot 12.4\,\mathrm{s} + 0.5\cdot (-5.14\,\mathrm{m/s^2})(12.4\,\mathrm{s})^2 ≈ 787.44\,\mathrm{m}$$ So, the jet stands about \(787.44\,\mathrm{m}\) down from its touchdown on the runway.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 train traveling at \(40.0 \mathrm{~m} / \mathrm{s}\) is headed straight toward another train, which is at rest on the same track. The moving train decelerates at \(6.0 \mathrm{~m} / \mathrm{s}^{2},\) and the stationary train is \(100.0 \mathrm{~m}\) away. How far from the stationary train will the moving train be when it comes to a stop?

An object starts from rest and has an acceleration given by \(a=B t^{2}-\frac{1}{2} C t,\) where \(B=2.0 \mathrm{~m} / \mathrm{s}^{4}\) and \(C=-4.0 \mathrm{~m} / \mathrm{s}^{3}\). a) What is the object's velocity after 5.0 s? b) How far has the object moved after \(t=5.0\) s?

A car moving at \(60.0 \mathrm{~km} / \mathrm{h}\) comes to a stop in \(t=4.00 \mathrm{~s}\) Assume uniform deceleration. a) How far does the car travel while stopping? b) What is its deceleration?

An electron moves in the positive \(x\) -direction a distance of \(2.42 \mathrm{~m}\) in \(2.91 \cdot 10^{-8} \mathrm{~s}\), bounces off a moving proton, and then moves in the opposite direction a distance of \(1.69 \mathrm{~m}\) in \(3.43 \cdot 10^{-8} \mathrm{~s}\). a) What is the average velocity of the electron over the entire time interval? b) What is the average speed of the electron over the entire time interval?

The trajectory of an object is given by the equation $$ x(t)=(4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) t-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} $$ a) For which time \(t\) is the displacement \(x(t)\) at its maximum? b) What is this maximum value?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Electricity

Read Explanation

Magnetism

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electromagnetism

Read Explanation

Medical 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.

Sign-up for free