Find the distance she walks

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.

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The Attempt at a Solution

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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Mentor

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.

??

The Attempt at a Solution

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.
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

Mentor

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?

Mentor

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