A jogger travels a route that has two parts. The first displacement A of 2.10 km

Question

A jogger travels a route that has two parts. The first displacement A of 2.10 km due south, and the second involves a displacement B that points due east.

a.) The resultant displacement A +B has a magnitude of 3.82 km. What is the magnitude of B? ________km

What is direction of A + B relative to due south? _________ degrees.

b.) Suppose A - B had a magnitude of 3.82 km. What would the magnitude of B and what is the direction of A - B relative to due south ? _____km

Direction of A - B relative to due south? ______degrees

east of south or west of south

A jogger travels a route that has two parts. The first is a displacement A with arrow of 2.10 km due south, and the second involves a displacement B with arrow that points due east.

(a) The resultant displacement A with arrow + B with arrow has a magnitude of 3.82 km. What is the magnitude of B with arrow?

km

What is the direction of A with arrow + B with arrow relative to due south?

°   

(b) Suppose that A with arrow - B with arrow had a magnitude of 3.82 km. What then would be the magnitude of B with arrow and what is the direction of A with arrow - B with arrow relative to due south?

magnitude of B with arrow

km

direction of A with arrow - B with arrow relative to due south

°

Explanation / Answer

a) distance due south A = 2.1 km

distance due east B = x

total displacement = 3.82 km

A+B = sqrt[ A^2+ B^2 + 2*A*B*cos(theta) ]

here the angle between A and B is 90 and cos 90 = 0

so A+B = sqrt( A^2+B^2 )

Total displacement = sqrt(A^2+B^2)

3.82 = sqrt(2.1^2+x^2)

x = 3.19 km

direction of A+b due south = arctan(B/A)

theta = arctan(3.19/2.1)

theta = 56.65 degrees

b) A - B = sqrt[ A^2 + B^2 - 2A*B*cos(tehta) ]

here the angle between A and B is 90 and cos 90 = 0

so A - B = sqrt( A^2+B^2 )

3.82 = sqrt(2.1^2 + B^2)

B = 3.19 km (due west since direction of B is opposite)

direction theta = arctan(-B/A)

theta = arctan(-3.19/2.1) = -56.65 degrees

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