Problem 47 - Algebra and Trigonometry

Question

Velocity of a Boat \(A\) straight river flows east at a speed of 10 milh. A boater starts at the south shore of the river and heads in a direction \(60^{\circ}\) from the shore (see the figure). The motorboat has a speed of 20 mi/h relative to the water. (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. (c) Find the true velocity of the motorboat. (d) Find the true speed and direction of the motorboat.

Step-by-Step Solution

Answer

True velocity: 28.76 mi/h at 20.47° north of east.

Step 1: Express River Velocity as a Vector

The river flows directly east at 10 mi/h. In vector components, this can be expressed as \( \vec{v}_r = \langle 10, 0 \rangle \). This means 10 units in the east (or positive x-direction) and 0 units in the north (y-direction).

Step 2: Express Boat Velocity Relative to Water

The boater heads in a direction \(60^{\circ}\) from the south shore, which makes a \(30^{\circ}\) angle with the x-axis (east). The speed of the boat is 20 mi/h relative to the water. In component form, it can be calculated using trigonometry:\[\vec{v}_b = 20 \cdot \langle \cos(30^{\circ}), \sin(30^{\circ}) \rangle = 20 \cdot \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle = \langle 10\sqrt{3}, 10 \rangle\]

Step 3: Calculate the True Velocity of the Motorboat

To find the true velocity, you need to sum the vector of the motorboat relative to the water and the river's velocity vector:\[\vec{v}_{\text{true}} = \vec{v}_b + \vec{v}_r = \langle 10\sqrt{3}, 10 \rangle + \langle 10, 0 \rangle = \langle 10\sqrt{3} + 10, 10 \rangle\]

Step 4: Determine the True Speed and Direction

The true speed of the boat can be found using the magnitude of the true velocity vector:\[|\vec{v}_{\text{true}}| = \sqrt{(10\sqrt{3} + 10)^2 + 10^2}\]The direction (angle \(\theta\) from the east) can be found by:\[\theta = \tan^{-1}\left(\frac{10}{10\sqrt{3} + 10}\right)\]

Step 5: Compute Numerical Values

Calculate the magnitude and direction numerically. The magnitude evaluates to:\[|\vec{v}_{\text{true}}| = \sqrt{(27.32)^2 + 10^2} \approx 28.76 \text{ mi/h}\]The direction is:\[\theta = \tan^{-1} \left(\frac{10}{27.32} \right) \approx 20.47^{\circ} \text{ north of east}\]

Key Concepts

velocity vectors

Velocity vectors are a fundamental concept in physics that allows us to describe the speed and direction of moving objects. In the context of vector calculus, we express velocity as a vector, meaning it has both magnitude (how fast an object is moving) and direction (where the object is heading). For instance, a velocity vector might say that an object is moving 10 miles per hour to the east.
In the exercise concerning the boater and the river, we express the velocities of both the river and the boat in vector form. This makes it easier to understand how each movement directionally affects the other.

• The river's velocity is straightforward as it's moving east at a constant speed. Therefore, its velocity vector components are east (x-axis) = 10, north (y-axis) = 0.
• The boat's velocity relative to the water is a bit more complex because it moves at an angle. Therefore, we use trigonometry to break it down into components.

Adding the two vectors gives us the true velocity of the motorboat, accounting for both its intended path and the current of the river.

trigonometry

Trigonometry is crucial for converting velocities at angles into vector components. In this exercise, the motorboat is heading 60 degrees from the south shore. To express this direction as a vector, we first determine how this angle relates to a standard coordinate system where east is along the positive x-axis.We use the trigonometric functions cosine and sine:

• The **cosine** of an angle in a right triangle describes the adjacent side over the hypotenuse, which in the case of a direction, corresponds to the horizontal (x-axis) component.
• The **sine** describes the opposite side over the hypotenuse, corresponding to the vertical (y-axis) component.

To find the boat's velocity vector, we multiply its speed (20 mi/h) with these cosine and sine values for the adjusted angle (30 degrees to the east in our exercise), yielding:For x-component (along east):\( v_x = 20 \cdot \cos(30^{\circ}) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \)For y-component (up north):\( v_y = 20 \cdot \sin(30^{\circ}) = 20 \cdot \frac{1}{2} = 10 \)

components of vectors

Understanding the components of a vector is like breaking down a puzzle into smaller pieces to solve it easier. A vector's components tell us precisely how much influence a vector has along each axis or direction, usually the x-axis (horizontal) and y-axis (vertical).By dissecting a vector into these components, we can perform vector addition easily, as vectors behave similarly to numbers along their respective axes.In the exercise, we break down the velocity vectors of both the river and the boat into their respective components:

• **River's velocity vector component:** Since it flows east directly, its components are simple: eastwards: 10, northwards: 0.
• **Boat's velocity vector component relative to water:** Here, using trigonometry, we decompose it into \(\text{x-axis: } 10\sqrt{3}\) (east) and \(\text{y-axis: } 10\) (north).

Adding these components gives the overall or true velocity vector of the boat, which in turn can be utilized to calculate the true speed and direction using vector mathematics, such as using the Pythagorean theorem for magnitude and inverse tangent for direction.