Question

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.

Answer 1

First segment displacement is given as

`d_1 = 260 km\: at \: 30^o \: noth \: of \: east`

so this is given in component form as

`d_1 = 260 cos30 \hat i + 260 sin30 \hat j`

`d_1 = 225.2 km \hat i + 130 km \hat j`

similarly the other displacement is given as

d2 = 176 km due west

`d_2 = -176 km \hat i`

now the total displacement is given as

`d = d_1 + d_2`

`d = 225.2 \hat i + 130 \hat j - 176 \hat i`

`d = 49.2 \hat i + 130 \hat j`

part a)

magnitude of the displacement is given as

`d = √(49.2^2 + 130^2)`

`d = 139 km`

part b)

for the direction of displacement we can say

`\theta = tan^(-1)(49.2)/(130)[tex]</p><p>[tex]\theta = 20.73 degree`

so it is 20.73 degree North of East