Problem 83 - University Physics with Modern Physics

Question

While following a treasure map, you start at an old oak tree. You first walk 825 \(\mathrm{m}\) directly south, then turn and walk 1.25 \(\mathrm{km}\) at \(30.0^{\circ}\) west of north, and finally walk 1.00 \(\mathrm{km}\) at \(40.0^{\circ}\) north of east, where you find the treasure: a biography of Isaac Newton! (a) To return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem. (b) To see whether your calculation in part (a) is reasonable, check it with a graphical solution drawn roughly to scale.

Step-by-Step Solution

Answer

Walk approximately 666 meters at 59.7° south of west.

Step 1: Convert Units

First, convert all distances to meters for consistency. The initial distance is already in meters: 825 meters. The second distance is 1.25 km, which equals 1250 meters, and the third distance is 1.00 km, which is 1000 meters.

Step 2: Convert Directions to Cartesian Components

Determine the component vectors for each leg of the journey. The walk directly south (\(v_1\)) has components: \(x_1 = 0\,\mathrm{m}\), \(y_1 = -825\,\mathrm{m}\). For the walk 1250 meters at 30.0° west of north (\(v_2\)), use trigonometry to get: \(x_2 = 1250\cos(150^{\circ})\), \(y_2 = 1250\sin(150^{\circ})\). For the walk 1000 meters at 40.0° north of east (\(v_3\)), use: \(x_3 = 1000\cos(40^{\circ})\), \(y_3 = 1000\sin(40^{\circ})\).

Step 3: Calculate Resultant Components

Add up all the x-components and y-components separately. \(R_x = x_1 + x_2 + x_3\) and \(R_y = y_1 + y_2 + y_3\).

Step 4: Calculate Distance to Return

Use the Pythagorean theorem to find the resultant displacement's magnitude: \(R = \sqrt{R_x^2 + R_y^2}\).

Step 5: Determine Direction to Return

Find the angle of the resultant vector with the x-axis using the inverse tangent function: \( \theta = \arctan\left(\frac{R_y}{R_x}\right) \). Ensure you take into account the quadrant in which the vector lies.

Step 6: Graphical Solution Check

Draw a rough scale diagram of the journey's path. Use a protractor and a ruler to represent each leg with appropriate angles and lengths, checking visually if the resultant vector leads back to the starting point.

Key Concepts

Trigonometry

Trigonometry is fundamental in understanding and solving vector problems, especially involving directions and angles. In this exercise, trigonometric functions such as sine and cosine are essential to convert directional vectors into Cartesian components. For example, while walking 1250 meters at 30.0° west of north, trigonometry helps break this path into x and y components using \( \cos(150^\circ) \) for the x-axis and \( \sin(150^\circ) \) for the y-axis.
In the trigonometric sense:

• Sine (\

Pythagorean theorem

The Pythagorean theorem is a mathematical principle used to determine the magnitude of a resultant vector. When you have two perpendicular components, as in our vector problem, the theorem makes it easy to find the length of the hypotenuse, which in this case, represents the total displacement from the starting point.
The formula is straightforward: \ R = \sqrt{R_x^2 + R_y^2} \ where \( R_x \) and \( R_y \) are the sums of all x and y directional changes respectively. This gives the shortest path back to the starting point. Understanding this can help demystify vector magnitudes and orientation. By calculating the hypotenuse, we know exactly how far we need to travel to return to the old oak tree.
It's a crucial tool in converting multi-leg trips to a direct path.

• Identify x and y components
• Apply the Pythagorean theorem to find the resultant
• Provides exact distance for return journey

Inverse tangent

The inverse tangent function, or arctan, is crucial for determining the angle of a resultant vector relative to a reference axis. In this case, it helps in finding out the direction you should head to return to the starting point. The function is given by: \[ \theta = \arctan\left(\frac{R_y}{R_x}\right) \] This calculates the angle theta, taking into consideration the vector's placement in a specific quadrant of the coordinate system.
Understanding the quadrant is key because:

• Vectors can be in different directions
• Helps identify positive and negative component values
• Ensures correct angle orientation

Ensuring the correct angle allows for precise navigation back to the initial position, helping to visualize directions on a compass. It's perfect for plotting accurate return paths in any vector-based navigation challenge.

Coordinate conversion

Coordinate conversion involves transitioning from polar coordinates, which are defined by a distance and an angle, to Cartesian coordinates, which are defined by x and y components on a grid. This conversion is imperative in vector mathematics and is necessary for problem-solving, like the path back to the oak tree. Each leg of the journey needs to be converted into Cartesian components to simplify calculations.
For example, if you walk 1.25 km at 30.0° west of north, you'd compute:

• Polar: Magnitude = 1.25 km, Angle = 30.0° west of north
• Convert to Cartesian: \( x = 1250\cos(150^\circ) \, y = 1250\sin(150^\circ) \)

Understanding this conversion helps assess x and y displacements, facilitating further calculations like summing components and determining resultant paths. This method is the backbone of solving multi-step navigation problems using vectors.