Steve S. answered • 02/01/14

Tutor

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Tutoring in Precalculus, Trig, and Differential Calculus

"The vector -5.2A has a magnitude of 47m and points in the positive x direction. Find the x component of the vector A."

"The vector -5.2A ... points in the positive x direction":

-5.2A = a u + 0 v; where u is the unit vector in the x-direction and v in y-direction (we don't have good symbols).

The magnitude of -5.2A, |-5.2A| = √(a^2 + 0^2) = a = 47

-5.2A = 47 u + 0 v

A = (47/-5.2) u + 0 v

"Find the x component of the vector A."

47/-5.2 = -9.03846153846154