Problem 12 - Fundamentals of Physics

Question

A car is driven east for a distance of \(50 \mathrm{~km}\), then north for 30 \(\mathrm{km}\), and then in a direction \(30^{\circ}\) east of north for \(25 \mathrm{~km}\). Sketch the vector diagram and determine (a) the magnitude and (b) the angle of the car's total displacement from its starting point.

Step-by-Step Solution

Answer

Magnitude: 80.5 km; Angle: 39.3° north of east.

Step 1: Draw the Vector Diagram

Draw a coordinate system with the starting point at the origin. The first movement is 50 km east, represented by a vector to the right along the x-axis. Next, from the endpoint of the first vector, draw a vector 30 km upwards along the y-axis (north). Finally, draw a vector of 25 km at 30° east of north, starting from the endpoint of the second vector. This vector can be broken into its northern and eastern components.

Step 2: Calculate Components of the Third Displacement

For the vector of 25 km at 30° east of north, use trigonometry to find the components. The northward component is calculated as \(25 \times \cos(30^{\circ}) = 21.65 \mathrm{~km}\), and the eastward component is \(25 \times \sin(30^{\circ}) = 12.5 \mathrm{~km}\).

Step 3: Determine Total Eastward and Northward Displacements

Add up all the eastward displacements: 50 km + 12.5 km = 62.5 km. Add up all the northward displacements: 30 km + 21.65 km = 51.65 km. These are the total displacements in the x (east) and y (north) directions.

Step 4: Calculate the Magnitude of the Total Displacement

Use the Pythagorean theorem to find the magnitude of the total displacement. The formula is \(\sqrt{(62.5)^2 + (51.65)^2}\). Compute this to get approximately 80.5 km.

Step 5: Calculate the Angle of Displacement

Use the tangent function to find the angle \(\theta\) from the east direction. The formula is \(\tan(\theta) = \frac{51.65}{62.5}\). Calculate \(\theta = \arctan\left(\frac{51.65}{62.5}\right)\) to find the angle. This angle is approximately 39.3° north of east.

Key Concepts

Displacement

Displacement is a vector quantity that indicates how far and in what direction something has moved. Unlike distance, which is scalar, displacement takes both magnitude and direction into account. This means it measures the shortest path from the initial to the final position and involves direction. In our example, to find the total displacement, you would consider both the distance traveled and the direction, forming straight lines between points on the path the car took. The vector diagram we discussed is crucial for visualizing this:

• First, the car moves 50 km east. This is a straight-line movement in the positive x-direction.
• Then, it goes 30 km north, moving upward in the positive y-direction.
• Finally, the car moves 25 km at a 30-degrees angle east of north.

Each of these movements is a vector, and by adding them correctly, you determine the resultant vector of the car's journey, which is its total displacement.

Trigonometry

Trigonometry offers us tools to analyze angles and lengths in right triangles. It becomes handy when calculating components of vectors that are not aligned with the coordinate axes. For the car problem, consider the final leg of the journey: a movement 25 km at 30° east of north. To use trigonometry here, we break down this movement into components. The northward (vertical) component is found using the cosine function, while the eastward (horizontal) component is determined using the sine function.

• For northward: o Use the formula: o \(North = 25 \times \cos(30^{\circ})\) This gives a value of approximately 21.65 km.
• For eastward: o Use the formula: o \(East = 25 \times \sin(30^{\circ})\) This results in 12.5 km.

Breaking vectors into components allows us to stick to the axis and add up parallel directions effectively. This process is essential for calculating net displacement.

Coordinate System

A coordinate system helps locate every point in a 2D or 3D space using axes. In this scenario, we use a two-dimensional Cartesian coordinate system, which defines locations with x (east-west) and y (north-south) coordinates. By placing the car's starting position at the origin, setting coordinates makes tracking its movement straightforward:

• The first vector, moving 50 km east, alters the position along the x-axis.
• The second vector, moving 30 km north, extends the position up the y-axis.
• The third vector, moving 25 km at an angle, requires breaking into parts that adjust both the x and y positions.

Using coordinates simplifies calculations and adjunct concepts like displacements, making them clearer and more manageable.

Pythagorean Theorem

The Pythagorean theorem is particularly useful for calculating the magnitude of a resultant vector when you know its horizontal and vertical components. It applies to right triangles, and it provides the hypotenuse, which in this case represents the total displacement of the car. The formula is given by \[C = \sqrt{A^2 + B^2}\], where C is the hypotenuse, and A and B are the other two sides. This offers a mathematical approach to determining vector magnitude. In the exercise:

• Total eastward displacement (A) is 62.5 km.
• Total northward displacement (B) is 51.65 km.

Substitute these values into the theorem: \[C = \sqrt{(62.5)^2 + (51.65)^2}\], resulting in a magnitude of approximately 80.5 km. This ensures we calculate the straight-line distance from start to end, representing the shortest path the car could have taken with those constraints.