The magnitude of the net displacement is 95.3 m
Explanation:
To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.
1st displacement is:
[tex]d_1=79 m[/tex] at [tex]16.9^{\circ}[/tex]
So its components are
[tex]d_{1x}=(79)(cos 16.9^{\circ})=75.6 m\\d_{1y}=(79)(sin 16.9^{\circ})=23.0 m[/tex]
2nd displacement is:
[tex]d_2=16.7 m[/tex] at [tex]31.1^{\circ}[/tex]
So its components are
[tex]d_{2x}=(16.7)(cos 31.1^{\circ})=14.3 m\\d_{2y}=(16.7)(sin 31.1^{\circ})=8.6 m[/tex]
Therefore, the x- and y-components of the net displacement are:
[tex]d_x=d_{1x}+d_{2x}=75.6+14.3=89.9 m\\d_y=d_{1y}+d_{2y}=23.0+8.6=31.6 m[/tex]
Therefore, the magnitude of the final displacement is:
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(89.9)^2+(31.6)^2}=95.3 m[/tex]
Learn more about displacement:
brainly.com/question/3969582
#LearnwithBrainly
