Eb6 What is the plane’s total displacement

[karush]

An airplane trip involves three legs, with two stopovers,
The first leg is due east for 620 km;
the second leg is south- east (\(\displaystyle 45^\circ\)) for 440 km
and the third leg is at 53\(\displaystyle ^\circ\) south of west, for 550 km
a. the first thing is to see if there is a way to graph this either with Desmos or someother online grapher
b. What is the plane’s total displacement
ok I know that we need to calculate its magnitude as well as direction for a complete discription.

 

[Prove It]

An airplane trip involves three legs, with two stopovers,
The first leg is due east for 620 km;
the second leg is south- east (\(\displaystyle 45^\circ\)) for 440 km
and the third leg is at 53\(\displaystyle ^\circ\) south of west, for 550 km
a. the first thing is to see if there is a way to graph this either with Desmos or someother online grapher
b. What is the plane’s total displacement
ok I know that we need to calculate its magnitude as well as direction for a complete discription.

Why bother trying to use technology to graph it when it's so easy to draw by hand?

Also, $53^{\circ}$ south of west, do you mean $S\,53^{\circ}\,W$ or $S\,37^{\circ}\,W$?

 

[HallsofIvy]

Prove It, "53 degrees S of W" is, properly, "S 37 degrees W". "S 53 degrees W" would be "53 degrees W of S".

Karush, The simplest thing to do is to set up a "coordinate system" and calculate the components of the vectors. Set up a coordinate system with the initial point as origin, (0, 0), positive x-axis to the right (east) and positive y-axis upward (north).

"An airplane trip involves three legs, with two stopovers,
The first leg is due east for 620 km;"
That would be to the right so we are now at (620, 0)

"The second leg is south- east ([FONT=MathJax_Main]45[/FONT][FONT=MathJax_Main]∘[/FONT]) for 440 km"
The second leg has components (440 cos(-45), 440 sin(-45))= (220sqrt(2), -220sqrt(2)) so we are now at (620+ 220sqrt(2). -220 sqrt(2))= (931, -311).

"and the third leg is at 53[FONT=MathJax_Main]∘[/FONT] south of west, for 550 km
"53 degrees south of west" is 180- 53= 127 degrees from the positive x-axis (east) so this motion would be (550 cos(127), 550 sin(127))= (-331, 439). We wind up at (931- 331. -311+ 439)= (600, 128).

"a. the first thing is to see if there is a way to graph this either with Desmos or someother online grapher"
Its really just a matter of drawing straight lines and measuring angles. Do have a protractor?
(Does anyone today know what a protractor is or have they gone the way or the slide rule?)

"b. What is the plane’s total displacement"
ok I know that we need to calculate its magnitude as well as direction for a complete discription."

The magnitude is sqrt(600^2+ 128^2) and the angle, measured from east, is 180+ arctan(128/600).

 

[karush]

Why bother trying to use technology to graph it when it's so easy to draw by hand?

Also, $53^{\circ}$ south of west, do you mean $S\,53^{\circ}\,W$ or $S\,37^{\circ}\,W$?

because I am making a pdf of these problemsand the third leg is at $53^\circ$

 

[Prove It]

Prove It, "53 degrees S of W" is, properly, "S 37 degrees W". "S 53 degrees W" would be "53 degrees W of S".

and the third leg is at $53^\circ$

I had a feeling, hence why I asked...

 

[HallsofIvy]

You had a feeling that Karush didn't know what it meant?

 

[karush]

View attachment 7672
ok think this is the basic idea