Question

An adventurous dog strays from home, runs a distance L at an angle θ north

of East, then walk a distance D directly south
a.) How far from home is the dog? Express your answer in terms of L, D, and θ
b.)In what direction is the dog?Express your angle with respect to east.

Answer 1

Step-by-step explanation:

To solve this problem, we will use vector addition. We can represent the dog's displacement as two vectors: one that points north of east, and another that points directly south. Let's call the first vector A and the second vector B.

a.) The distance from home is the magnitude of the total displacement vector, which we can find by adding vectors A and B using the Pythagorean theorem:

|A + B| = sqrt[(L cosθ + (-D))^2 + (L sinθ)^2]

Simplifying this expression gives:

|A + B| = sqrt[L^2 + D^2 - 2LD cosθ]

Therefore, the distance from home is sqrt[L^2 + D^2 - 2LD cosθ].

b.) To find the direction of the displacement, we need to find the angle between the displacement vector and the eastward direction. We can use trigonometry to do this:

tanθ = opposite/adjacent = L sinθ / (L cosθ - D)

Simplifying this expression gives:

tanθ = (sinθ)/(cosθ - D/L)

Taking the inverse tangent of both sides gives:

θ = tan^-1[(L sinθ)/(L cosθ - D)]

We can express this angle with respect to east by subtracting 90 degrees if θ is acute (i.e., θ < 90 degrees) or adding 90 degrees if θ is obtuse (i.e., θ > 90 degrees).