A jogger travels a route that has two parts. The first is a displacement of A-->

Question

A jogger travels a route that has two parts. The first is a displacement of A-->1.90 km due south, and the second involves a displacement B--> that points due east. The resultant displacement A--> + B--> has a magnitude of 4.40 km. (a) What is the magnitude of B--> , and what is the direction of A--> + B--> as a positive angle relative to due south? (b) Suppose that A--> - B--> had a magnitude of 4.40 km. What then would be the magnitude of B-->, and what is the direction of A--> - B--> relative to due south?

Explanation / Answer

A=-1.9 j

B= B i

a). |A+B|=4.4

1.9^2 + B^2 =4.4^2

B=3.97

thus

B=3.97 i

thus A+B=3.97i -1.9 j

angle=tan-1(1.9/3.97)

=25.58 to down from +ve x axis

b).|A-B|=4.4

1.9^2 + B^2 =4.4^2

B=3.97

thus

B=3.97i

A-B=-1.9j - 3.97 i

thus angle=25.58 to down from -ve x axis

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