Jetliner Displacement: Magnitude & Direction in Northwest - 66*, 27*

  • Thread starter Thread starter possum30540
  • Start date Start date
  • Tags Tags
    Vector
Click For Summary

Homework Help Overview

The problem involves a jetliner's displacement after takeoff, requiring the calculation of both the magnitude and direction of its displacement in a three-dimensional context. The subject area includes vector analysis and trigonometry.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the construction of triangles to determine angles related to the displacement. There are questions about the appropriate use of components and whether the horizontal component is always considered the Z component. Some participants suggest alternative approaches to defining the triangles used for calculations.

Discussion Status

The discussion is ongoing, with participants offering clarifications and corrections to each other's reasoning. Some have expressed understanding after reviewing their approaches, while others are still questioning the setup of their calculations.

Contextual Notes

There is a mention of potential confusion regarding the definitions of components and the correct application of trigonometric principles in the context of the problem. Participants are also reflecting on their previous knowledge and the need for accuracy in their calculations.

possum30540
Messages
17
Reaction score
1

Homework Statement



A jetliner takes off from LA. after 5 min, the jet has an altitude of 5 km above the level of the airport and is 25 km west and 10 km north of the airport. What are the magnitude and the diretion of its displacement during this time interval? Specify the direction by stating two angles: the angle of the displacements direction west of north and the angle of the displacement's direction above the horizontal.

Homework Equations



for West of North:
construct a triangle with adjacent side 10 km and hypotenuse 25 km
then use cox=10/25 to find the angle

For above the horizontal:
construct a triangle with opposite side 5 km and hypotenuse 10
then use tanx=5/10 to find angle

For magnitude:

use the magnitude formula for: [-25 km, 10 km, 5 km]


The Attempt at a Solution



Is horizontal always the Z component/up?

west of north angle: 66*
above horizontal angle: 27*
direction of displacement: northwest
magnitude: 27 km
 
Physics news on Phys.org
possum30540 said:
for West of North:
construct a triangle with adjacent side 10 km and hypotenuse 25 km
then use cox=10/25 to find the angle

For above the horizontal:
construct a triangle with opposite side 5 km and hypotenuse 10
then use tanx=5/10 to find angle

It's been awhile since I've done any of this so somebody correct me if otherwise, but for west of north and for above the horizontal, you should not be using any distance as the hypotenuse.

You will need two different triangles for each situation which you did, but the hypotenuse is not known until you calculate it but you don't need it.
 
Okay . .I understand now that I look at it again. Thank you.

for north of west:
construct triangle with opposite 25 and adjacent 10

for above horizontal:
construct triangle with opposite 5 and adjacent 10
 
I believe so yes, that is what I had written down I think. I erased it but I think the angles I had were something like 68.198* and 26.565*
 

Similar threads

Replies
10
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
Replies
2
Views
5K
  • · Replies 19 ·
Replies
19
Views
8K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K