What Is the Total Distance Walked by the Woman in This Vector Problem?

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Homework Help Overview

The problem involves a woman walking a total distance with specified directions, requiring the calculation of her final displacement magnitude and angle, as well as the total distance walked. The subject area pertains to vector analysis and trigonometry.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of vector components based on angles measured from different axes. There is an exploration of the correct interpretation of the angle given in the problem, with some questioning the use of angles from the positive x-axis versus the y-axis.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the assumptions made regarding angle measurements. Some participants express confusion about the calculations and the angle conversions, while others clarify their approaches and results.

Contextual Notes

There is a noted difference in how angles are interpreted, which may affect the calculations. Participants are also grappling with the distinction between total distance walked and displacement.

sagaradeath
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Homework Statement



A woman walks 132 m in the direction 41° east of north, then 170 m directly east. Find (a) the magnitude and (b) the angle (from due east) of her final displacement from the starting point. (c) Find the distance she walks.

Homework Equations





??

The Attempt at a Solution



i think it is wrong please help

N displacement of leg 1 = (cos41) x 132m., = 99.62m.
E displacement of leg 1 = (sin 41) x 132m., = 86.6m.

Total N displacement = 99.62m.
Total E displacement = (170 + 86.6) = 256.6m.

a) Magnitude = sqrt. (99.62^2 + 256.6^2), = 275.26 m.
b) Sin L = (99.62/275.26), so L = 22.22 degrees, N of E.
c) (132 + 170) = 302 m.
 
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sagaradeath said:

Homework Statement



A woman walks 132 m in the direction 41° east of north, then 170 m directly east. Find (a) the magnitude and (b) the angle (from due east) of her final displacement from the starting point. (c) Find the distance she walks.

Homework Equations





??

The Attempt at a Solution



i think it is wrong please help

N displacement of leg 1 = (cos41) x 132m., = 99.62m.
E displacement of leg 1 = (sin 41) x 132m., = 86.6m.
41 deg. E of North is an angle of 49 deg counterclockwise from the positive x-axis. Use 49 deg in your calculations above.
sagaradeath said:
Total N displacement = 99.62m.
Total E displacement = (170 + 86.6) = 256.6m.

a) Magnitude = sqrt. (99.62^2 + 256.6^2), = 275.26 m.
b) Sin L = (99.62/275.26), so L = 22.22 degrees, N of E.
c) (132 + 170) = 302 m.
 


wait I don't understand
 


What don't you understand? When you're calculating the components of vectors, the angles need to be measured from the pos. x-axis. The heading 41 deg. E of North is measured from the y-axis.
 


so it would be like this

sin(49) x 132=99.62?
 


That would be the vertical component of the first leg of her walk.
 


yeah i don't get sorry :(
 


You are doing your calculations with an angle measured from the y-axis. I'm doing calculations with an angle measured from the x-axis. We're both getting the same answers for the vertical component (N displacement) and horizontal component (E displacement).

I get a slightly different angle, using the N and E displacements and arctan, getting 21.2 deg N of E.
 

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