Physics word problem

Question

A jogger travels a route that has two parts. The first is a displacement Á of 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 3.90 km. (a) What is the magnitude of B, and (b) what is the direction of A + B as a positive angle relative to due south? Suppose that Ä - B had a magnitude of 3.90 km. (c) What then would be the magnitude of B, and (d) what is the direction of Ä - B relative to due south?

Answer 1

Part a

We know displacement A is due south and displacement B is due east, so the two displacements are perpendicular to each other. Since there is a 90 degree angle between the displacement, the given two displacements and their sum must form a right triangle. So we can use the Pythagorean theorem. Let C = A + B be the hypotenuse, then:

A² + B² = C²

We know A and C, so solve for B

B = sqrt(C² - A²) = sqrt((3.90 km)² - (1.90 km)²)

B = 3.41 km

Part b

The angle relative to due south has an adjacent side of A and a hypotenuse of C, so

cos(θ) = A/C

or

θ = cos^-1(A/C) = cos^-1( (1.9 km) / (3.9 km)) θ

θ = 60.8 degrees

Part c

Define D = A - B

but A still points due south and B due east. . Note that A - B will point southwest instead of southeast.

This still forms a right triangle but the hypotenuse and leg A still have the same length as part a, so B is also the same length

Part d

Since B has the same magnitude the angle is the same value, but now points to the southwest, which we could label as -60.8 degrees or add 360 degrees to call it positive +299.2 degrees