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Margaret walks to the store using the following path: 0.850 mi west, 0.390 mi north, 0.150 mi east. Assume north to be along the +y-axis and west to be along the –x-axis.Find the direction of the displacement vector. Enter the answer as an angle north of west

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In order to find the displacement vector, first let's add the vertical components (north being positive and south being negative) and the horizontal components (east being positive and west being negative):


\begin{gathered} \text{vertical:} \\ 0.39 \\ \\ \text{horizontal:} \\ 0.15-0.85=-0.7 \end{gathered}

the horizontal component is negative and the vertical component is positive, so the direction is towards Quadrant II.

To find the angle, first let's calculate the arc tangent of the quotient between the vertical and horizontal components:


\begin{gathered} \theta=\tan ^(-1)((0.39)/(-0.7)) \\ \theta=\tan ^(-1)(-0.55714) \\ \theta=150.88\degree \end{gathered}

Now, to find the angle north of west, let's find the supplementary angle of theta:


180-150.88=29.12\degree

Margaret walks to the store using the following path: 0.850 mi west, 0.390 mi north-example-1
User Yan Koshelev
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