Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 33: Problem 13

An airplane makes a round trip between two points \(1800 \mathrm{km}\) apart, flying with airspeed \(800 \mathrm{km} / \mathrm{h}\). What's the round trip flying time (a) if there's no wind, (b) with wind at 130 km/h perpendicular to a line joining the two points, and (c) with wind at \(130 \mathrm{km} / \mathrm{h}\) along a line joining the two points?

Short Answer

Expert verified
The round trip flying times are (a) 4.5 hours, (b) 4.5 hours, and (c) 4.61 hours respectively.

Step by step solution

01

Calculate Flying Time with No Wind

Start by calculating the flying time with no wind, using the formula flying time = distance / speed. Thus, the round trip flying time will be \(2 \times (1800 \, \mathrm{km} / 800 \, \mathrm{km/h}) = 4.5 \, \mathrm{h}\).
02

Calculate Flying Time with Wind Perpendicular

Next, calculate the flying time when the wind is perpendicular to the line joining plane positions; in this case, the wind speed has no effect on the airplane's velocity. Thus, the round trip flying time is still \(4.5 \, \mathrm{h}\) as in part (a).
03

Calculate Flying Time with Wind Along

Finally, calculate the flying time when the wind is along the line joining plane positions. For the onward journey, the wind increases the airplane's speed, whereas for the return path, the wind reduces it. The total round trip flying time equals the sum of the onward and return flying times. Thus, the time for onward journey = distance / (airspeed + wind speed) = \(1800 \, \mathrm{km} / (800+130) \, \mathrm{km/h} = 1.88 \, \mathrm{h}\), and the time for return journey = distance / (airspeed - wind speed) = \(1800 \, \mathrm{km} / (800-130) \, \mathrm{km/h} = 2.73 \, \mathrm{h}\). Adding these two times gives the total round trip flying time = \(1.88 + 2.73 = 4.61 \, \mathrm{h}\).

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

Radioactive oxygen-15 decays at such a rate that half the atoms in a given sample decay every 2 min. If a tube containing 1000 O-15 atoms is moved at \(0.80 c\) relative to Earth for 6.67 min according to clocks on Earth, how many atoms will be left at the end of that time?

A light beam is emitted at event A and arrives at event B. Show that the spacetime interval between the two events is zero.

A spaceship travels at \(0.80 c\) from Earth to a star 10 light years distant, as measured in the Earth-star reference frame. Let event A be the ship's departure from Earth and event B its arrival at the star. (a) Find the distance and time between the two events in the Earth-star frame. (b) Repeat for the ship's frame. (Hint: The distance in the ship frame is the distance an observer has to move with respect to that frame to be at both events-not the same as the Lorentz-contracted distance between Earth and star.) (c) Compute the square of the spacetime interval in both frames to show explicitly that it's invariant.

Compare the momentum changes needed to boost a spacecraft (a) from 0.1 \(c\) to \(0.2 c\) and (b) from 0.8 \(c\) to \(0.9 c .\)

Is matter converted to energy in a nuclear reactor? In a burning candle? In your body?
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Fields in Physics

Read Explanation

Linear Momentum

Read Explanation

Atoms and Radioactivity

Read Explanation

Solid State Physics

Read Explanation

Dynamics

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