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

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The discussion revolves around calculating the total distance walked by a woman who first walks 132 m at an angle of 41° east of north and then 170 m directly east. The calculations for the north and east displacements are provided, with the north component calculated as approximately 99.62 m and the east component as about 256.6 m. The final displacement magnitude is determined to be approximately 275.26 m, with an angle of about 22.22° north of east. There is a clarification regarding the angle measurement, emphasizing the need to measure angles from the positive x-axis rather than the y-axis. The total distance walked is confirmed to be 302 m.
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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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