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Q1.

Expert-verifiedFound in: Page 67

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**A car is driven 225 km west and then 98 km southwest (45°). What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.**

** **

The magnitude of the displacement of the car from the point of origin is 302.3 km, and the direction is $13.2\xb0$south of the west.

**The net displacement of an object can be defined as the vector addition of the object's individual displacement in various directions. If the object's initial and final points are similar, the object's net displacement value is zero.**

Given data:

** **

The displacement of the car in the western direction is ${\overrightarrow{D}}_{west}=225\mathrm{km}$.

The displacement of the car in the southwestern direction $\left(\theta =45\xb0\right)$is ${\overrightarrow{D}}_{south-west}=98\mathrm{km}$.

The vector diagram for the problem can be drawn as follows:

The resultant displacement vector in the western direction can be calculated as

$\begin{array}{c}{\overrightarrow{D}}_{R-west}={\overrightarrow{D}}_{west}+{\overrightarrow{D}}_{south-west}\mathrm{cos}\theta \\ {\overrightarrow{D}}_{R-west}=\left(225\mathrm{km}\right)+\left(98\mathrm{km}\right)\mathrm{cos}\left(45\xb0\right)\\ {\overrightarrow{D}}_{R-west}=294.3\mathrm{km}\end{array}$

The resultant displacement vector in the southwestern direction can be calculated as

$\begin{array}{c}{\overrightarrow{D}}_{R-southwest}={\overrightarrow{D}}_{south-west}\mathrm{sin}\theta \\ {\overrightarrow{D}}_{R-southwest}=\left(98\mathrm{km}\right)\mathrm{sin}\left(45\xb0\right)\\ {\overrightarrow{D}}_{R-southwest}=69.3\mathrm{km}\end{array}$

The magnitude of the resultant displacement vector can be calculated as

$\begin{array}{c}{\overrightarrow{D}}_{R}=\sqrt{{\left({\overrightarrow{D}}_{R-west}\right)}^{2}+{\left({\overrightarrow{D}}_{R-southwest}\right)}^{2}}\\ {\overrightarrow{D}}_{R}=\sqrt{{\left(294.3\mathrm{km}\right)}^{2}+{\left(69.3\mathrm{km}\right)}^{2}}\\ {\overrightarrow{D}}_{R}=302.3\mathrm{km}\end{array}$

The direction for the resultant displacement vector can be calculated as

$\begin{array}{c}\theta ={\mathrm{tan}}^{-1}\left(\frac{{\overrightarrow{D}}_{R-southwest}}{{\overrightarrow{D}}_{R-west}}\right)\\ \theta ={\mathrm{tan}}^{-1}\left(\frac{69.3\mathrm{km}}{294.3\mathrm{km}}\right)\\ \theta =13.2\xb0\end{array}$

Thus, the magnitude of the displacement of the car from the point of origin is 302.3 km, and the direction is $13.2\xb0$south of the west.

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