Question

A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 235 km and a direction 30.0^o north of east. The displacement vector B for the second segment has a magnitude of 173 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

Suppose given that Vector is R1 and it makes angle with +x-axis, then it's components are given by:

R1x = R1*cos

R1y = R1*sin

Using above rule:

A = 235 km, direction 30.0 deg N of E

Ax = 235*cos 30 deg = 203.516 km

Ay = 235*sin 30 deg = 117.5 km

B = 173 km, direction towards west

Bx = 173*cos 180 deg = -173 km

By = 173*sin 180 deg = 0 km

Now resultant displacement will be:

R = A + B

R = (Ax + Bx) i + (Ay + By) j

R = (203.516 + (-173)) i + (117.5 + 0) j

R = 30.516 i + 117.5 j

So distance R will be

|R| = sqrt (Rx^2 + Ry^2)

|R| = sqrt (30.516^2 + 117.5^2)

|R| = 121.4 km = magnitude of R

Direction will be given by:

= arctan (Ry/Rx)

= arctan (117.5/30.516)

= 75.4 deg North of east

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