Vector $\vec { A }$ has a magnitude of 6.0 m and points 30° north of east. Vector $\overrightarrow { \mathbf { B } }$ has a magnitude of 4.0 m and points 30° east of north. The resultant vector $\vec { A }$ + $\overrightarrow { \mathbf { B } }$ is given by
A) 0.70 m at an angle of 42° north of east.
B) 14 m at an angle of 42° north of east.
C) 1.1 m at an angle of 42° north of east.
D) 9.7 m at an angle of 42° north of east.
E) 2.0 m at an angle of 42° north of east.

Question

Quizplus · Accepted Answer

To find the resultant vector, we add the x-components and y-components of the two vectors separately.

The x-component of vector A is A*cos(30°) = 6*cos(30°) = 5.20 m.
The y-component of vector A is A*sin(30°) = 6*sin(30°) = 3.00 m.

The x-component of vector B is B*sin(30°) = 4*sin(30°) = 2.00 m.
The y-component of vector B is B*cos(30°) = 4*cos(30°) = 3.46 m.

Adding the x-components gives us a resultant x-component of 5.20 + 2.00 = 7.20 m.
Adding the y-components gives us a resultant y-component of 3.00 + 3.46 = 6.46 m.

Using the Pythagorean theorem, we find the magnitude of the resultant vector:
|R| = sqrt((7.20)^2 + (6.46)^2) = 9.7 m.

To find the direction of the resultant vector, we use the inverse tangent function:
tan^(-1)(y-component / x-component) = tan^(-1)(6.46 / 7.20) = 42° north of east.

Therefore, the answer is D, 9.7 m at an angle of 42° north of east.

Vector A⃗\vec { A }A Has a Magnitude of 6 B→\overrightarrow { \mathbf { B } }B

Vector $\vec { A }$ Has a Magnitude of 6 $\overrightarrow { \mathbf { B } }$