Let’s solve each part systematically.

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

The river flows east at a speed of $11 \, \text{mi/h}$. Since the river's flow is entirely in the east ($\mathbf{i}$) direction: $\mathbf{v}_\text{river} = 11 \mathbf{i} + 0 \mathbf{j}$

So: $\mathbf{v}_\text{river} = (11, 0) \, \text{mi/h}.$

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(b) Express the velocity of the motorboat relative to the water as a vector in component form.

The motorboat's speed is $20 \, \text{mi/h}$, and it is headed at an angle of $60^\circ$ from the shore (or horizontal). Using trigonometric components:

$\mathbf{v}_\text{boat-water} = \text{speed} \times (\cos\theta \mathbf{i} + \sin\theta \mathbf{j}),$

where: $\text{speed} = 20, \quad \theta = 60^\circ.$

Substitute values: $\mathbf{v}_\text{boat-water} = 20 \cos(60^\circ) \mathbf{i} + 20 \sin(60^\circ) \mathbf{j}.$

Using $\cos(60^\circ) = 0.5$ and $\sin(60^\circ) = 0.866$: $\mathbf{v}_\text{boat-water} = 20(0.5) \mathbf{i} + 20(0.866) \mathbf{j}.$

$\mathbf{v}_\text{boat-water} = 10 \mathbf{i} + 17.32 \mathbf{j}.$

So: $\mathbf{v}_\text{boat-water} = (10, 17.32) \, \text{mi/h}.$

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(c) Find the true velocity of the motorboat.

The true velocity is the vector sum of the river's velocity and the motorboat's velocity relative to the water: $\mathbf{v}_\text{true} = \mathbf{v}_\text{river} + \mathbf{v}_\text{boat-water}.$

Substitute the vectors: $\mathbf{v}_\text{true} = (11, 0) + (10, 17.32).$

Add the components: $\mathbf{v}_\text{true} = (11 + 10, 0 + 17.32).$

$\mathbf{v}_\text{true} = (21, 17.32) \, \text{mi/h}.$

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(d) Find the true speed of the motorboat.

The true speed is the magnitude of the true velocity vector: $\text{Speed} = \sqrt{v_x^2 + v_y^2},$

where $v_x = 21$ and $v_y = 17.32$: $\text{Speed} = \sqrt{21^2 + 17.32^2}.$

$\text{Speed} = \sqrt{441 + 300.29}.$

$\text{Speed} = \sqrt{741.29} \approx 27.2 \, \text{mi/h}.$

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(e) Find the direction of the motorboat (angle with respect to the east).

The direction angle $\theta$ is given by: $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right).$

Substitute $v_x = 21$ and $v_y = 17.32$: $\theta = \tan^{-1}\left(\frac{17.32}{21}\right).$

$\theta \approx \tan^{-1}(0.824).$

$\theta \approx 39.2^\circ.$

So the direction is: $39.2^\circ \, \text{N of E}.$

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Final Answers

(a) $\mathbf{v}_\text{river} = (11, 0) \, \text{mi/h}$

(b) $\mathbf{v}_\text{boat-water} = (10, 17.32) \, \text{mi/h}$

(c) $\mathbf{v}_\text{true} = (21, 17.32) \, \text{mi/h}$

(d) True speed = $27.2 \, \text{mi/h}$

(e) Direction = $39.2^\circ \, \text{N of E}$

Let me know if you want further details or clarifications! Here are some related questions to explore:

1. How would the motorboat's speed change if the angle was $45^\circ$ instead of $60^\circ$?
2. What happens to the true velocity if the river's speed increases to $15 \, \text{mi/h}$?
3. How would the problem differ if the river flowed west instead of east?
4. Can we generalize the true speed and direction for any angle and river speed?
5. What is the significance of the relative velocity concept in real-life scenarios like aviation or boating?

Tip: Visualizing velocity vectors graphically can make these problems much clearer!