How to Calculate Magnitude and Direction for Orv's Walk?

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Homework Help Overview

The problem involves calculating the magnitude and direction of Orv's walk after he initially walks 312 m due east and then stops 220 m northeast of his starting point. The context is rooted in vector analysis and trigonometry.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the Pythagorean theorem and trigonometric functions to determine the magnitude and direction of the second portion of Orv's walk. There are questions about the correct application of these concepts and the interpretation of the problem's parameters.

Discussion Status

Some participants have offered guidance on using diagrams and the correct relationships between the sides of the triangle formed by Orv's path. There is an ongoing exploration of the assumptions regarding the triangle's configuration and the relationships between the distances involved.

Contextual Notes

There are indications of confusion regarding the roles of the distances in the problem, particularly concerning which values represent the legs and hypotenuse of the triangle. Participants also note the importance of accurately representing the final position in a diagram.

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Homework Statement


Orv walks 312 m due east. He then continues walking along a straight line, but in a different direction, and stops 220 m northeast of his starting point. How far did he walk during the second portion of the trip and in what direction?
What is the magnitude?
What is the direction ___________ ° counterclockwise from the +x-axis



Homework Equations


C=squareroot of 312^2+220^2
What equation to use to find direction??

The Attempt at a Solution


C= Magnitude= 381.76 m why is my magnitude wrong?
 
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A good idea here is to read the problem carefully, and/or make a simple diagram.
You're correct in using the Pythagorean theorem for this problem, but you're using it incorrectly. Remember that the form is a^2 + b^2 = c^2, where c is the hypotenuse.

In the problem statement, you're given a = 312 meters (The distance walked east -- one of the legs of the triangle), and c = 220 meters (The distance in a straight line from the starting point -- the hypotenuse).

This means that what you need to find here is b, not c.
 
As with regards to your direction issues, once you have your three sides figured out, use your trig laws (soh cah toa).

If you've never heard of them, it goes:
"soh" - sin = opposite / hypotenuse
"cah" - cosine = adjacent / hypotenuse
"toa" - tangent = opposite / adjacent

And you should get the right answer.
Cheers

EDIT: or your sin/cosine laws. I didn't work through the problem myself, I just assumed based on the level it was a right angle triangle.
 
Gordanier said:
EDIT: or your sin/cosine laws. I didn't work through the problem myself, I just assumed based on the level it was a right angle triangle.

Same. Now that I look at it (through my sleepy eyes), the 'hypotenuse' is shorter than one of the legs, and by a good bit, too...
 
I can't take the square root of a negative number for the magnitude... Is there another way to solve this...
 
lalahelp said:

Homework Statement


Orv walks 312 m due east. He then continues walking along a straight line, but in a different direction, and stops 220 m northeast of his starting point. How far did he walk during the second portion of the trip and in what direction?
What is the magnitude?
What is the direction ___________ ° counterclockwise from the +x-axis

Did you ever draw it?
 

Attachments

  • Triangle16Sep11.jpg
    Triangle16Sep11.jpg
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@:AC130NAV: The problem states that the final position was 220 m from the starting point in an north easterly direction, so your drawing is not correct.
 
Last edited:

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