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A dog looking for its bone in the yard, first heads 160.0° for 14.0 m and then turns and heads west for 8.50 m.

a. What is the magnitude of the dog's displacement?
b. What is the direction of the dog's displacement?

1 Answer

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To find the magnitude of the dog's displacement and its direction, we can use vector addition. We'll break this problem into two parts:

1. The dog heads 160.0° for 14.0 m. Let's call this vector A.
2. Then, the dog turns and heads west for 8.50 m. This vector points directly west, so it's along the negative x-axis. Let's call this vector B.

a. To find the magnitude of the dog's displacement, we'll use the Pythagorean theorem because these vectors form a right triangle:

Displacement magnitude = √((A^2) + (B^2))

Where:
A = 14.0 m
B = -8.50 m (negative because it's west)

Plugging in the values:

Displacement magnitude = √((14.0^2) + (-8.50^2))
Displacement magnitude ≈ √(196 + 72.25)
Displacement magnitude ≈ √268.25
Displacement magnitude ≈ 16.37 meters (rounded to two decimal places)

b. To find the direction of the dog's displacement, we can use trigonometry. The angle θ (theta) can be found using the inverse tangent (arctan) function:

θ = arctan(B / A)

Where:
A = 14.0 m
B = -8.50 m (negative because it's west)

Plugging in the values:

θ = arctan(-8.50 / 14.0)
θ ≈ -30.51°

Since the dog initially headed at 160.0°, and then turned west (270.0°), the final direction can be found by adding these angles:

Final direction = 160.0° + (-30.51°) = 129.49°

So, the magnitude of the dog's displacement is approximately 16.37 meters, and its direction is approximately 129.49° (measured counterclockwise from the east direction).
User Vijay S B
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