Vector A has a magnitude of 40.0 m and points in the direction of 20.0 degrees b

Question

Vector A has a magnitude of 40.0 m and points in the direction of 20.0 degrees below the +x axis. Vector B has a magnitude of 75.0 m and points in the direction of 50.0 degrees above the +x axis. Before doing the following calculations, first draw each vector. Also use the Graphical Method (meaning making drawings) to do vector addition and vector subtraction. Review lecture notes and examples, and get help if you are not sure how to do Graphical Method.

Part A - Vector C = Vector A + Vector B, calculate the magnitude of Vector C. Use 3 digits.

Part B - Vector C = Vector A + Vector B, calculate the direction of Vector C. Report the direction by a counterclockwise angle from the +x axis. Use 3 digits.

Part C - Vector D = Vector A - Vector B, calculate the magnitude of Vector D.  Use 3 digits.

Part D - Vector D = Vector A - Vector B, calculate the direction of Vector D. Report the direction by a counterclockwise angle from the +x axis. Use 4 digits.

Explanation / Answer

Given,

A = 40 m ; theta = 20 deg

B = 50 m ; theta = 50 deg

Ax = 40 x cos20 = (37.588 m) i

Ay = 40 x sin(-20) = (-13.681 m) j

A = 37.588 i - 13.681 j

Bx = 50 x cos50 = (32.139 m )i

By = 50 x sin50 = (38.302 m)j

B = 32.139 i + 38.302 j

A) C = A + B

C = 37.588 i - 13.681j + 32.139 i + 38.302 j = 69.726 i + 24.621 j

magnitude of C will be:

C = sqrt (69.726^2 + 24.621^2) = 73.945 m

Hence, lCl = 73.945 m

B)theta = tan^ -1 (24.621/69.726) = 19.449 deg

Hence, theta = 19.449 deg

C) D = A - B

D = 37.588 i - 13.681j - 32.139 i - 38.302 j = 5.449 i - 51.983 j

D = sqrt (5.449 + -51.983^2) = 52.268 m

Hence, D = 52.268 m

D)theta = tan^-1(-51.983/5.449) = -84.0159 deg

Hence, theta = -84.0159 deg

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