Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Construct Your Own Problem Consider an airplane headed for a runway in a cross wind. Construct a problem in which you calculate the angle the airplane must fly relative to the air mass in order to have a velocity parallel to the runway. Among the things to consider are the direction of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air mass. Also calculate the speed of the airplane relative to the ground. Discuss any last minute maneuvers the pilot might have to perform in order for the plane to land with its wheels pointing straight down the runway.

Short Answer

Expert verified
  • 69.3ms
  • 81.8โˆ˜.

Step by step solution

01

Definition of velocity

Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.

The velocity of the plane relative to the earth will be considered resultant.

This problem is the relative velocity problem; we can use the relative velocity formula.

02

Velocity of the plane relative to earth

Hence the velocity can be calculated by the following equation.

Vpe=Vpa+Vae

The component table will be as below:

X

Y

Vpa

-70cosฮธ

-70sinฮธ

Vae

10

0

Resultant

0 (plane is landing parallel to runway)

-70sinฮธ

Here we need to find the x component and the y component.

XcomponentVpax=VicosฮธVpax=-70cosฮธYcomponentVpay=VisinฮธVpay=-70sinฮธ

Hence, solving the resultant equation to get the value of theta:

-70cosฮธ+10=070cosฮธ=10cosฮธ=1070ฮธ=81.8ยฐ.

The angle of the direction will be81.8ยฐ .

03

Determine the resultant velocity

Hence the resultant vector will be:

Rโ‡€=โˆ‘X2+โˆ‘Y2Rโ‡€=02+-70sin81.82Rโ‡€Vpe=63.3ms

The velocity of the plane relative to the earth is 63.3ms.

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 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

The world long jump record is8.95m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of9.5m/s? State your assumptions

An archer shoots an arrow at a 75.0mdistant target; the bullโ€™s-eye of the target is at same height as the release height of the arrow.

(a) At what angle must the arrow be released to hit the bullโ€™s-eye if its initial speed is 35.0m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems.

(b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch3.50m above the release height of the arrow. Will the arrow go over or under the branch?

The great astronomer Edwin Hubble discovered that all distant galaxies are receding from our Milky Way Galaxy with velocities proportional to their distances. It appears to an observer on the Earth that we are at the center of an expanding universe. The figure illustrates this for five galaxies lying along a straight line, with the Milky Way Galaxy at the center. Using the data from the figure, calculate the velocities:

(a) relative to galaxy 2

(b) relative to galaxy 5.

The results mean that observers on all galaxies will see themselves at the center of the expanding universe, and they would likely be aware of relative velocities, concluding that it is not possible to locate the center of expansion with the given information.

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50km45.0ยฐnorth of west; then 4.70km60.0ยฐsouth of east; then 1.30km25.0ยฐsouth of west; then 5.10kmstraight east; then 1.70km5.00ยฐeast of north; then 7.20km 55.0ยฐsouth of west; and finally 2.80km10.0ยฐnorth of east. What is his final position relative to the island?

Repeat Exercise using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same resultโ€”that is, B+A=A+B.) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path.

See all solutions

Recommended explanations on Physics Textbooks

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