A pilot flies her route in two straight-line segments. The displacement vector a for the first segment has a magnitude of 260 km and a direction 30.0o north of east. The displacement vector b for the second segment has a magnitude of 176 km and a direction due west. The resultant displacement vector is r = a + b and makes an angle θ with the direction due east. Using the component method, find (a) the magnitude of r and (b) the directional angle θ.

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aristocles aristocles · Answer 1 · 2018-09-21T03:42:43+02:00

First segment displacement is given as

[tex]d_1 = 260 km\: at \: 30^o \: noth \: of \: east[/tex]

so this is given in component form as

[tex]d_1 = 260 cos30 \hat i + 260 sin30 \hat j[/tex]

[tex]d_1 = 225.2 km \hat i + 130 km \hat j[/tex]

similarly the other displacement is given as

d2 = 176 km due west

[tex]d_2 = -176 km \hat i[/tex]

now the total displacement is given as

[tex]d = d_1 + d_2 [/tex]

[tex]d = 225.2 \hat i + 130 \hat j - 176 \hat i[/tex]

[tex]d = 49.2 \hat i + 130 \hat j[/tex]

part a)

magnitude of the displacement is given as

[tex]d = \sqrt{49.2^2 + 130^2}[/tex]

[tex]d = 139 km[/tex]

part b)

for the direction of displacement we can say

[tex]\theta = tan^{-1}\frac{49.2}{130}[tex]

[tex]\theta = 20.73 degree[/tex]

so it is 20.73 degree North of East