Answer:
(a) 82 Km
(b) 32°
Explanation:
First you should draw the vectors in the cartesian plane (please see the picture below).
As the car is driven east for a distance of 47 Km, your first vector should be drawn from the origin (0,0) and on the x axis
Then the car is driven north for a distance of 28 Km, so your second vector should be drawn from the origin and on the y axis.
And the car finally goes east of north for 27Km, so the third vector should be drawn from the origin east of north forming an 35° angle with x axis.
Then you should find the components of the vector in x and y:
For Vector 1 ([tex]V_{1}[/tex])
[tex]V_{1x}=47Km[/tex]
[tex]V_{1y}=0[/tex]
For Vector 2 ([tex]V_{2}[/tex])
[tex]V_{2x}=0[/tex]
[tex]V_{2y}=28Km[/tex]
For Vector 3 ([tex]V_{3}[/tex])
[tex]V_{3x}=27cos37^{o}[/tex]
[tex]V_{3x}=21.56[/tex]
[tex]V_{3y}=27sin37^{o}[/tex]
[tex]V_{3y}=16.25[/tex]
To find the magnitud of the car´s total displacement, (R) you should add up all the x and y components.
For the x component:
[tex]R_{x}=V_{1x}+V_{2x}+V_{3x}[/tex]
[tex]R_{x}=47+0+21.56[/tex]
[tex]R_{x}=68.56Km[/tex]
For the x component:
[tex]R_{y}=V_{1y}+V_{2y}+V_{3y}[/tex]
[tex]R_{y}=0+28+16.25[/tex]
[tex]R_{y}=44.25Km[/tex]
Now please see the second picture that is showing the components x and y as the sides of a right triangle, and we are going to use the Pythagorean theorem to find the resultant, R.
[tex]R=\sqrt{R_{x}^{2}+R_{y}^{2}[/tex]
[tex]R=\sqrt{68.56^{2}+44.25^{2}}[/tex]
[tex]R=82Km[/tex]
And to find the angle of the car´s total displacement (α), we use the same right triangle with the relationship between its legs.
[tex]tan(\alpha)=\frac{44.25}{68.56}[/tex]
[tex]tan(\alpha)=0.64[/tex]
[tex]\alpha=tan^{-1}(0.64)[/tex]
[tex]\alpha =32^{o}[/tex]
