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Expert-verified Found in: Page 57 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A person desires to reach a point that is${\mathbf{3}}{\mathbf{.}}{\mathbf{40}}{\mathbf{}}{\mathbf{km}}$from her present location and in a direction that is${\mathbf{3}}{\mathbf{.}}{\mathbf{50}}{\mathbf{°}}$north of east. However, she must travel along streets that are oriented either north–south or east–west. What is the minimum distance she could travel to reach her destination?

The minimum distance traveled to reach destination is$4.47\mathrm{km}$

See the step by step solution

## Step 1: To understand the concept of scalar projection

This problem refers to scalar projection. In Cartesian coordinates, scalar components are scalar projections in the directions of the coordinate axes. Using this concept, the distance can be calculated by finding the components and adding those components.

The components can be written as

${d}_{x}=d\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{d}_{y}=d\mathrm{sin}\theta$

Therefore the minimum distance D is given by the following formula.

$D={d}_{x}+{d}_{y}=d\mathrm{cos}\theta +d\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)$

## Step 2: To find the minimum distance traveled

Given are,

$d=3.40\mathrm{km}\phantom{\rule{0ex}{0ex}}\theta =35.0°$

Substituting the above values in equation (i), the minimum distance D can be written as,

$D=\left[3.40×\mathrm{cos}\left(35.0\right)\right]+\left[3.40×\mathrm{sin}\left(35.5\right)\right]$

Thus, $D=4.47\mathrm{km}$

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