Judy walks a certain distance due north, then walks twice as far due west. At the end of her trip, the straight-line distance from her starting point to her final position is 112m.
a) What is the length of each leg of the trip?
b) What is the direction of her total displacement relative to north?
Please help ASAP
x^2 + (2x)^2 = 112
The angle θ West of North is
tanθ = 2/1
Still need help
To solve this problem, we can break it down into two steps:
Step 1: Determine the lengths of each leg of the trip.
Let's assume that Judy walks a distance of x meters due north. Then, she walks twice that distance due west, which is 2x meters.
According to the problem, the straight-line distance from Judy's starting point to her final position is 112 meters. This straight-line distance forms a right triangle with the legs representing the distances traveled due north and due west.
Using the Pythagorean theorem, we can calculate the length of each leg:
Leg due north (x): since this distance is perpendicular to the leg due west, it forms the vertical side of the right triangle.
Leg due west (2x): since this distance is perpendicular to the leg due north, it forms the horizontal side of the right triangle.
Applying the Pythagorean theorem to the right triangle:
(x^2) + (2x^2) = 112^2
3x^2 = 112^2
x^2 = (112^2)/3
x^2 ≈ 4157.78
x ≈ √4157.78
x ≈ 64.45
Therefore, the length of the leg due north is approximately 64.45 meters, and the length of the leg due west is twice as long, so it is approximately 2 * 64.45 = 128.90 meters.
So, the length of each leg of Judy's trip is approximately 64.45 meters due north and 128.90 meters due west.
Step 2: Determine the direction of her total displacement relative to north.
To find the direction of her total displacement, we can use simple trigonometry. We can find the angle between the straight-line distance and due north.
In this case, the angle θ is the angle opposite to the leg due north.
Using the tangent function:
tan(θ) = (opposite/adjacent) = (128.90/64.45)
tan(θ) = 2
Now, we can find the angle θ by taking the inverse tangent (also known as arctangent) of 2.
θ = arctan(2)
θ ≈ 63.43°
Therefore, the direction of Judy's total displacement relative to north is approximately 63.43 degrees west of north.