Pre calculus HW

Question

A straight river flows east at a speed of 12 mi/h. A boater starts at the south shore of the river and heads in a direction 60° from the shore. The motorboat has a speed 24 mi/h relative to the water. (Assume that the i vector points east, and the j vector points north.)

(a) Express the velocity of the river as a vector in component form.

(b) Express the velocity of the motorboat relative to the water as a vector in component form. (Round your answer to two decimal places.)

(c) Find the true velocity of the motorboat. (Round your answer to two decimal places.)

(d) Find the true speed of the motorboat. (Round your answer to one decimal place.)

Find the direction of the motorboat. (Round your answer to one decimal place.)

Answer 1

(a) The velocity of the river can be expressed as a vector in component form as:

velocity of the river = 12i

since the river flows only in the east direction.

(b) The velocity of the motorboat relative to the water can be expressed as a vector in component form using trigonometry. Let's define the angle between the direction of the motorboat and the direction of the river as θ. Then, the horizontal and vertical components of the velocity of the motorboat relative to the water are:

1. Horizontal component: 24 cos(θ)
2. Vertical component: 24 sin(θ)

Since the angle between the motorboat and the river is 60°, we have:

cos(θ) = cos(60°) = 1/2 sin(θ) = sin(60°) = sqrt(3)/2

Substituting the values, we get:

1. Horizontal component: 24 * 1/2 = 12
2. Vertical component: 24 * sqrt(3)/2 ≈ 20.8

Therefore, the velocity of the motorboat relative to the water can be expressed as a vector in component form as:

velocity of the motorboat relative to the water = 12i + 20.8j

(c) The true velocity of the motorboat is the vector sum of the velocity of the motorboat relative to the water and the velocity of the river. We can add the components separately:

1. Horizontal component: 12 + 12 = 24
2. Vertical component: 20.8

Therefore, the true velocity of the motorboat can be expressed as a vector in component form as:

true velocity of the motorboat = 24i + 20.8j

(d) The true speed of the motorboat is the magnitude of the true velocity of the motorboat:

true speed of the motorboat = ||true velocity of the motorboat|| = sqrt(24^2 + 20.8^2) ≈ 31.1 mi/h

(e) The direction of the motorboat can be found using the tangent function:

tan(θ) = 20.8/24

Solving for θ, we get:

θ ≈ 39.5°

Therefore, the direction of the motorboat is approximately 39.5° north of east.