A pilot flies her route in two straight-line segments. The displacement vector A

Question

A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 253 km and a Erection 30.0 degree 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 theta with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle theta. (a) R = Number Units (b) theta = Number Units

Explanation / Answer

a] R = [253* cos 30 degree - 173] E + 253 sin 30 degree N

= 46.1 E + 126.5 N

= sqrt(46.1^2+126.5^2)

= 134.6 km

b] direction angle = arctan(126.5/46.1)

= 69.98 degree

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