A jogger travels a route that has two parts. The first is a displacement of 2.15 km due south, and the second involves a displacement that points due east. The resultant displacement + has a magnitude of 3.95 km. (a) What is the magnitude of , and what is the direction of + as a positive angle relative to due south? (b) Suppose that - had a magnitude of 3.95 km. What then would be the magnitude of , and what is the direction of - relative to due south?

Y = -2.15 km.

R = 3.95 km.

a. sin A = Y/R = -2.15/3.95 = -0.54430
A = -33o = 57o East of South.
R = 3.95km[57o] East of South.

To solve this problem, we can draw a diagram to represent the displacements and use the Pythagorean theorem and trigonometry to find the magnitudes and directions.

(a) Magnitude of the resultant displacement (+):
To find the magnitude of the resultant displacement (+), we can use the Pythagorean theorem because the two displacements are at right angles to each other. The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let's call the magnitude of the second displacement D. According to the problem, the magnitude of the first displacement is 2.15 km south, so it remains the same. The magnitude of the resultant displacement (+) is given as 3.95 km.

Therefore, we have the following equation:
(2.15 km)^2 + D^2 = (3.95 km)^2

Simplifying the equation, we have:
4.6225 + D^2 = 15.6025

Subtracting 4.6225 from both sides:
D^2 = 15.6025 - 4.6225
D^2 = 10.98

Taking the square root of both sides:
D ≈ 3.316 km

So, the magnitude of the second displacement is approximately 3.316 km.

To find the magnitude (+), we use the magnitude of the second displacement D and the magnitude of the resultant displacement (+):
Magnitude (+) = 2.15 km + 3.316 km ≈ 5.466 km

Hence, the magnitude of the resultant displacement (+) is approximately 5.466 km.

Direction of the resultant displacement (+):
To find the direction of the resultant displacement (+) as a positive angle relative to due south, we can use trigonometry. We can use the inverse tangent function to find the angle.

Let's call the angle we want to find θ.

Using trigonometry, we have:
tan(θ) = (2.15 km) / (3.316 km)

Taking the inverse tangent of both sides:
θ ≈ tan^(-1)(2.15 / 3.316)

Evaluating on a calculator, we find:
θ ≈ 34.81°

Therefore, the direction of the resultant displacement (+) is approximately 34.81° relative to due south.

(b) Magnitude of the resultant displacement (-):
If the magnitude of the resultant displacement (-) is given as 3.95 km, we can find the magnitude of the second displacement (-) using the same method as before.

Let's call the magnitude of the second displacement D'.

According to the problem, the magnitude of the first displacement remains the same as 2.15 km.

So, we have the following equation:
(2.15 km)^2 + (D')^2 = (3.95 km)^2

Simplifying the equation, we have:
4.6225 + (D')^2 = 15.6025

Subtracting 4.6225 from both sides:
(D')^2 = 15.6025 - 4.6225
(D')^2 = 10.98

Taking the square root of both sides:
D' ≈ 3.316 km

Therefore, the magnitude of the second displacement (-) is approximately 3.316 km.

To find the magnitude (-), we use the magnitude of the second displacement D' and the magnitude of the resultant displacement (-):
Magnitude (-) = 2.15 km - 3.316 km ≈ -1.166 km

Hence, the magnitude of the resultant displacement (-) is approximately -1.166 km (which means it is 1.166 km in the opposite direction of the initial displacement).

Direction of the resultant displacement (-) relative to due south:
Since the resultant displacement (-) is in the opposite direction of the initial displacement, the direction is 180° - 34.81° = 145.19° relative to due south.

Therefore, the direction of the resultant displacement (-) is approximately 145.19° relative to due south.