Problem 63 - Algebra and Trigonometry

Question

Velocity of a Boat A straight river flows east at a speed of 10 \(\mathrm{mi} / \mathrm{h}\) . 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 \(\mathrm{mi} / \mathrm{h}\) 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

The true velocity vector of the motorboat is \(\langle 10 + 10\sqrt{3}, 10 \rangle\), with a speed of approximately 27.32 mi/h and direction of around 18.43° north of east.

Step 1: River Velocity Vector

The river flows east at 10 mi/h. This direction corresponds to the positive x-axis in our coordinate system. Therefore, the velocity vector of the river, \(\mathbf{v}_r\), is expressed as \(\mathbf{v}_r = \langle 10, 0 \rangle\).

Step 2: Motorboat Velocity Relative to Water

The motorboat heads 60° from the south shore, which corresponds to 30° north of east. The speed of the motorboat relative to the water is 20 mi/h. We use these to find the components: the x-component (eastward) is \(20\cos(30^{\circ})\), and the y-component (northward) is \(20\sin(30^{\circ})\). Therefore, the velocity vector of the boat relative to the water, \(\mathbf{v}_m\), is \(\mathbf{v}_m = \langle 20\cos(30^{\circ}), 20\sin(30^{\circ}) \rangle\).

Step 3: Calculate Motorboat's Velocity Components Relative to Water

Calculate numerical values: \(\cos(30^{\circ}) = \sqrt{3}/2\) and \(\sin(30^{\circ}) = 1/2\). Therefore, \(\mathbf{v}_m = \langle 20(\sqrt{3}/2), 20(1/2) \rangle = \langle 10\sqrt{3}, 10 \rangle\).

Step 4: Determine True Velocity of Motorboat

The true velocity of the motorboat, \(\mathbf{v}_t\), is the sum of the river's velocity and the boat's velocity relative to the water. Thus, \(\mathbf{v}_t = \mathbf{v}_r + \mathbf{v}_m = \langle 10, 0 \rangle + \langle 10\sqrt{3}, 10 \rangle = \langle 10 + 10\sqrt{3}, 10 \rangle\).

Step 5: Calculate True Speed of the Motorboat

The true speed of the motorboat is the magnitude of the true velocity vector. This is calculated as \(\| \mathbf{v}_t \| = \sqrt{(10 + 10\sqrt{3})^2 + 10^2}\). Carry out the calculation to find the numeric value.

Step 6: Direction of the Motorboat

The direction \(\theta\) is obtained using the inverse tangent of the y-component divided by the x-component: \(\theta = \tan^{-1}\left(\frac{10}{10 + 10\sqrt{3}}\right)\). Calculate this to find the angle north of east.

Key Concepts

Trigonometry

Trigonometry plays a crucial role in breaking down the motion of objects when they are not moving strictly along the axes of a coordinate system. By using trigonometric functions like sine and cosine, we can decompose a vector into its horizontal and vertical components.
This is particularly useful when dealing with angles, as in the case of the motorboat heading at 60° from the south shore.

• Sine (\( ext{sin}\)) relates the opposite side of a triangle to its hypotenuse.
• Cosine (\( ext{cos}\)) relates the adjacent side to the hypotenuse.

Using these functions, we can calculate the x-component of the motorboat's velocity as \(20 ext{cos}(30^{\circ})\) and the y-component as \(20 ext{sin}(30^{\circ})\). Calculations like these make trigonometry a fundamental tool in physics to handle vectors.

Vector Components

Understanding vector components is essential for describing movements accurately in physics. A vector can be split into two or three components—depending on whether we're dealing with 2D or 3D problems—that align with the coordinate axes.
Components allow us to simplify complex directions into manageable parts.

• X-component: This represents movement along the x-axis, such as eastward or westward motion.
• Y-component: This represents movement along the y-axis, such as northward or southward motion.

When dealing with the motorboat's velocity, the vector component form helps us visualize how it progresses both in the eastward and northward directions. For instance, the boat's velocity relative to the water has components \(\langle 10\sqrt{3}, 10 \rangle\), providing a clear idea of its actual path.

Physics in Motion

Physics in motion often involves the study of how forces and movements interact, especially when different elements are involved, like in this case: the river flow and the motorboat. Understanding the true velocity of an object when it’s influenced by another moving medium is key.
To resolve this, we analyze the velocity vectors: the velocity of the river, \(\mathbf{v}_r = \langle 10, 0 \rangle\), and that of the boat relative to water. Both vectors need to be added to find the true velocity \(\mathbf{v}_t\).
This teaches us that motion is often relative and that true movement results from the vector sum of all influencing velocities. This fundamental principle underscores the complexity and beauty of physics in motion, providing deeper insights into how objects really move.

Coordinate Systems

Coordinate systems are vital tools for visualizing and solving problems involving motion, as they allow us to define directions and positions unambiguously. A common choice is the Cartesian coordinate system, with x and y axes representing east-west and north-south directions, respectively.
This simplification makes solving exercises like the velocity of the motorboat clearer. By assigning vectors to these axes, we can easily see how each component interacts:

• The river's velocity vector aligns perfectly along the x-axis, with no y-component.
• The motorboat's velocity is initially given in terms of an angle, which we transform into vector components by utilizing the coordinate system and trigonometric relationships.

Coordinate systems thus not only organize how we solve problems but also help in visualizing the real-world scenarios, easing the complexity of motion descriptions.