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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# You are to make four straight-line moves over a flat desert floor, starting at the origin of an xy coordinate system and ending at the xy coordinates (-140 m,30 m . The x component and y component of your moves are the following, respectively, in meters:(20 and 60) , then $\left({b}_{x}and-70\right)$ , then (-20 ${{\mathbf{ands}}}_{{\mathbf{y}}}$) , then (-60 and -70) . What are (a) component and (b) component ? What are (c) the magnitude and (d) the angle (relative to the positive direction of the x axis) of the overall displacement?

a) Component ${\mathrm{b}}_{\mathrm{x}}\mathrm{is}-80\mathrm{m}$

b) Component ${\mathrm{c}}_{\mathrm{y}}\mathrm{is}110\mathrm{m}$

c) Magnitude of the overall displacement is 140 m

d) Angle of the overall displacement is $-168°$

See the step by step solution

## Step 1: To understand the concept of Vectors

The vectors can be added in the same direction and equate that to the component of the resultant vector in the same direction. This would give the missing values of the components.

To evaluate the missing components, the following formulae can be used.

$\stackrel{⇀}{a}+\stackrel{⇀}{b}+\stackrel{⇀}{c}+\stackrel{⇀}{d}=\left({a}_{x}+{b}_{x}+{c}_{x}+{d}_{x}\right)\stackrel{⇀}{\mathrm{i}}+\left({a}_{y}+{b}_{y}+{c}_{y}+{d}_{y}\right)\stackrel{⇀}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}R=\sqrt{{R}_{X}^{2}+{R}_{y}^{2}}$ (i)

$\mathrm{Given}\phantom{\rule{0ex}{0ex}}\mathrm{R}=\left(-140,30\right)\phantom{\rule{0ex}{0ex}}\mathrm{A}=\left(20,60\right)\phantom{\rule{0ex}{0ex}}\mathrm{B}=\left({\mathrm{b}}_{\mathrm{x}},-70\right)\phantom{\rule{0ex}{0ex}}\mathrm{C}=\left(-20,{\mathrm{c}}_{\mathrm{y}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{D}=\left(-60,70\right)$ (ii)

## Step 2: To find component bx

To find ${b}_{x}$we have to add all x components of all vectors which is equal to-140 because components of resultant are (-140m,30m)

$20+{b}_{x}-20-60=-140\phantom{\rule{0ex}{0ex}}{b}_{x}=-80\mathrm{m}$

## Step 3: To find component cy

Now to find ${c}_{y}$ add all y components of all which is equal to 30 because components of resultant are(-140m,30m)

$60+{c}_{y}-70-70=30\phantom{\rule{0ex}{0ex}}{c}_{y}=110\mathrm{m}$

## Step 4: To find magnitude of resultant

As components of resultant are (-140,30) , so resultant is given as

$R=\sqrt{{\left(-140\right)}^{2}+{\left(30\right)}^{2}}\phantom{\rule{0ex}{0ex}}R=143\mathrm{m}$

## Step 5: To find direction of resultant,

$\mathrm{tan}\theta =\frac{30}{-140}\phantom{\rule{0ex}{0ex}}\theta =-12°$

The angle is negative which means it is measured in clockwise direction with respect to negative x axis. This can be also concluded from the fact that the x component of resultant is negative and y component is positive. So resultant is in the second quadrant. With respect to positive x axis angle would be $180°-12°=168°$ .

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