Ruth sets out to visit her friend Ward, who lives 50 mi north and 100 mi east of her. She starts by driving east, but after 60 mi she comes to a detour that takes her 15 mi south before going east again. She then drives east for 8 mi and runs out of gas, so Ward flies there in his small plane to get her.
Part A
What is Ward's displacement vector? Give your answer in component form, using a coordinate system in which the y-axis points north.
Express your answers using two significant figures separated by a comma.
rx,ry = mi
Part B
What is the magnitude of Ward's displacement vector?
Express your answer to two significant figures and include the appropriate units.
Part C
What is the direction of Ward's displacement vector, measured clockwise from the negative y axis?
Express your answer using two significant figures.

Because she is moving closer to the positive x-axis in 60 miles after which she also flied east in 8 miles. Also, she went south in 15 miles which is down the negative y-axis.

Thus, displacement vector is;

x = x2 - x1 = 68i - 15j - (100i + 50j)

x = 68i - 15j - 100i - 50j

x = -32i - 65j

In component form, we have;

x = (-32, - 65)

B) Magnitude of wards displacement vector;

|x| = √((-32)² + (-65²))

|x| = √(1024 + 4225)

|x| = √5249

|x| ≈ 72 miles

C) direction of Ward's displacement vector, measured clockwise from the negative y axis is;

θ = tan^(-1) (-32/-65)

θ = tan^(-1) 0.4923

θ = 26°

oeerivona oeerivona · Answer 2 · 2022-01-30T21:19:50+01:00

The given parameters of the location of Ruth and Ward can be

presented on a coordinate plane.

Correct response;

Part A; <-32, -65>

Part B; Approximately 72 miles

Part C; Approximately 26°

Method of calculation

Taking the initial location of Ruth as the origin, we have;

The initial location of Ward in component form = <100, 50>

The distances Ruth traveled in vector form are;

60·i - 15·j + 8·i = 68·i - 15·j

In component form, the distance Ruth traveled = <68, -15>

Ward's displacement vector is therefore;

rₓ = 68 - 100 = -32

[tex]r_y[/tex] = -15 - 50 = -65

Which gives;

Ward's displacement vector in component form, is; [tex]\vec d[/tex] = <-32, -65>

Part B

The magnitude of Ward's displacement vector, [tex]|\vec d |[/tex] is found as follows;

[tex]| \vec d |[/tex] = [tex]\sqrt{32^2 + 65^2} = \sqrt{5249} \approx \mathbf{72}[/tex]

The magnitude of Ward's displacement vector is [tex]|\vec d | \approx \underline{72 \ miles}[/tex][tex]{}[/tex]

Part C

The unit vector form of Ward's displacement = -32·i - 65·j

Therefore, the direction, ∅, with respect to a counterclockwise rotation from the negative x-axis is given as follows;

[tex]The \ direction \ \phi = arctan\left(\dfrac{-65}{-32} \right) \approx \mathbf{ 64^{\circ}}[/tex]

Which gives, the direction, θ, with respect to the negative y-axis is therefore;

θ ≈ 90° - 64° = 26°

The direction of Ward's displacement vector measured clockwise from the negative y-axis is approximately 26°

Learn more about the component form of a vector here:

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