Answer:
[tex]\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m[/tex]
Explanation:
Sum of Vectors in the Plane
Given two vectors
[tex]\displaystyle \vec{v_1}\ ,\ \vec{v_2}[/tex]
They can be expressed in their rectangular components as
[tex]\displaystyle \vec{v_1}=<x_1\ ,\ y_1>[/tex]
[tex]\displaystyle \vec{v_2}=<x_2\ ,\ y_2>[/tex]
The sum of both vectors can be done by adding individually its components
[tex]\displaystyle \vec{v_1}+\vec{v_2}=<x_1+x_2\ ,\ y_1+y_2>[/tex]
If the vectors are given as a magnitude and an angle [tex](M\ ,\ \theta )[/tex], each component can be found as
[tex]\displaystyle \vec{v_1}=<M_1 cos\theta_1\ ,\ M_1sin\theta _1>[/tex]
[tex]\displaystyle \vec{v_2}=<M_2 cos\theta_2\ ,\ M_2sin\theta_2>[/tex]
The first vector has a magnitude of 3.14 m and an angle of 30°, so
[tex]\displaystyle \vec{v_1}=<3.14\ cos30^o,3.14\ sin30^o>[/tex]
[tex]\displaystyle \vec{v_1}=<2.72,1.57>[/tex]
The second vector has a magnitude of 2.71 m and an angle of -60°, so
[tex]\displaystyle \vec{v_2}=<2.71cos(-60^o),2.71sin(-60^o)>[/tex]
[tex]\displaystyle \vec{v_2}=<1.36,-2.35>[/tex]
The sum of the vectors is
[tex]\displaystyle \vec{v_1}+\vec{v_2}=<2.72+1.36,1.57-2.35>[/tex]
[tex]\displaystyle \vec{v_1}-\vec{v_2}=<4.08,-0.78>[/tex]
Finally, we compute the magnitude of the sum
[tex]\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{(4.08)^2+(-0.78)^2}[/tex]
[tex]\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{17.25}[/tex]
[tex]\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m[/tex]
