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Q20 P

Expert-verifiedFound in: Page 59

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**A girl delivering newspapers covers her route by traveling $3.00$** **blocks west, $6.00$** **blocks north, and then $6.00$****blocks east. **

**(a) What is her resultant displacement? **

**(****b) What is the total distance she travels?**

a) The resultant displacement of the girl is 5 blocks at $53.1\xb0$ (North of East).

b) The total distance travelled by the girl is 13 blocks.

Displacement of a body is defined as the body moving from its rest position to another one. Distance is defined as the length of space between two points.

To distribute newspapers, the girl begins at point A and follows the course ABCD.

The resultant displacement

$\text{AO}=\sqrt{{\text{AD}}^{2}+{\text{OD}}^{2}}$

Now substitute $4.0$ for AD and $3.0$ for OD.

$\text{AO}=\sqrt{{4.0}^{2}+{3.0}^{2}}\phantom{\rule{0ex}{0ex}}=5\text{}$

Calculate the angle $\theta $ :

$\mathrm{tan}\theta =\frac{\text{AD}}{\text{OD}}$

Substitute 4.0 for AD and 3.0 for OD.

$\mathrm{tan}\theta =\frac{4}{3}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}\theta =53.1\xb0\text{(North of East)}$

Here, $\theta $is the direction.

Hence, the resultant displacement of the girl is 5 blocks at $53.1\xb0$ (North of East).

The total distance travelled.

$\text{D}=\text{AB}+\text{BC}+\text{CO}$

Substitute 3.0 for AB, 4.0 for BC and 6.0 for CO.

$\text{D}=3.0+4.0+6.0\phantom{\rule{0ex}{0ex}}=13\text{blocks}$

Hence, the total distance travelled by the girl is 13 blocks.

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