Vector A has a magnitude of 7.60 units and makes an angle of 44.0° counter-clock

Question

Vector A has a magnitude of 7.60 units and makes an angle of 44.0° counter-clockwise from the positive x-axis. Vector B has a magnitude of 8.00 units and is directed along the negative x-axis.

(a) Using graphical methods, find the vector sum A + B.

Magnitude of A + B:    

                 units

Direction of A + B:

                 ° counterclockwise from +x-axis

(b) Using graphical methods, find the vector difference A - B.

Magnitude of A - B:

                         units

Direction of A - B:

                        ° counterclockwise from +x-axis

Magnitude of A + B:    

                 units

Direction of A + B:

                 ° counterclockwise from +x-axis

Explanation / Answer

x component of vector A=7.6*cos(44)=5.822 units

y component of vector A=7.6*sin(44)=5.2794 units

in vector notation, A=5.822 i + 5.2794 j

x component of vector B=-8 units

y component of vector B=0

in vector notation, B=-8 i

part a:

vector A+B=5.822 i + 5.2794 j -8 i=-2.178 i + 5.2794 j

magnitude =sqrt(2.178^2+5.2794^2)=5.711 units

angle with +ve x axis=arctan(5.2794/(-2.178))=112.418 degrees

part b:

vector A-B=5.822 i + 5.2794 j + 8i=13.822 i + 5.2794 j

magnitude=sqrt(13.822^2+5.2794^2)=14.7959 units

angle with +ve x axis=arctan(5.2794/13.822)=20.9 degrees counter clockwise from x axis

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