Having a little trouble, very typical question

[DB]

im not use to working is 3D so i need some help

you are going to a restaurant. you leave your apartement and take the elevator 30 m down, then 15 m south to the exit. then u travel 200 m east and then 100 m to the restaurant. what is your displacement.

so what i did is:

-30k
-15j
+200i
+100j

then after adding them up:

\Delta d=\sqrt{200i^2+85j^2+(-30k^2)}

displacement = ~ 219.37 m

is this right? and how do i represent the final direction?

thanks

 

[quasar987]

You might want to double check Pythagora's thm there! It says\Delta d=\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}

 

[DB]

sry, i can't find the mistake your talking about, would that 85 be -115?

 

[Farsight]

That's an interesting one DB. I just checked the arithmetic and you do have the right answer in terms of distance. But whilst I can know how to use tangents to work out the horizontal direction in degrees that you ought to supply in addition to the distance, I'm not familiar with using this to express the downward component. Perhaps one of the more learned chaps can chip in here.

 

[quasar987]

you are going to a restaurant. you leave your apartement and take the elevator 30 m down, then 15 m south to the exit. then u travel 200 m east and then 100 m to the restaurant. what is your displacement.

Firstly, how do we know in which direction the last 100m is?

 

[quasar987]

sry, i can't find the mistake your talking about, would that 85 be -115?

I was referring to the fact that you did not square the distances. I.e. you used the equation

\Delta d = \sqrt{\Delta x+\Delta y+\Delta z}

Edit: Oh i understand! By 200i², you mean (200i)². The parenthesis are important because 200i^2 means (200)*(i²)=200.

 

[DB]

oh I am sorry, it is suposed to say 100m North

 

[HallsofIvy]

you are going to a restaurant. you leave your apartement and take the elevator 30 m down, then 15 m south to the exit. then u travel 200 m east and then 100 m (NORTH)to the restaurant. what is your displacement.]

I found myself not certain as to whether "displacement" was a vector (my first choice) or just the distance (which several of the answers are assuming) so I went to "google" to check. No luck! What I got was "A vector or the magnitude of a vector from the initial position to the final position"! In other words, it could be either and I think you had better check exactly which meaning it is given in your class.

In any case: taking the "i" vector to the east, the "j" vector north, the "k" vector upward, you: "take the elevator 30 m down": -30k; "then 15 m south to the exit": -15j;"then u travel 200 m east": 200i; "then 100 m (NORTH)to the restaurant" (North added from your later respons): 100j.
Adding all those, your displacement vector is -30k- 15j+ 200i+ 100j= 200i+ 85j- 30k. Your numerical displacement (distance) is
\sqrt{(200)^2+ (85)^2+ (-30)^2}= \sqrt{40000+ 7225+ 900}= \sqrt{48125} m.

 

[Farsight]

You got the distance portion right. To get the direction you need to know use tangents. You know that you moved 85m North and 200m East. You can express these as an opposite over adjacent ratio, and find the bearing in degrees using the INV tangent function on this here online calculator.

http://www.math.com/students/calculators/source/scientific.htm

You can then repeat with 30/sqrt(85² + 200²) to get a declination.

But I suspect that the answer they were looking for was just the 219.37m distance.

 

[DB]

thanks everyone, btw hallsofivy displacement is a vector, distance is not. just to cleat that up :) thanks again i understand the problem now