Question

A man in search of his dog drives first 10 mi northeast, then \(12 \mathrm{mi}\) straight south, and finally \(8 \mathrm{mi}\) in a direction \(30^{\circ}\) north of west. What are the magnitude and direction of his resultant displacement?

Short answer

Answer: The man's resultant displacement has a magnitude of approximately 3.75 mi and is at an angle of 33.1° from the eastward direction (to the northeast).

Step-by-step solution

Break down the directions into components

First, we need to break down each direction into their individual components (East-North or West-South). Remember that East and North have positive values, while West and South have negative values. Trip 1: 10 mi northeast is 10 mi East and 10 mi North. East component: 10 mi North component: 10 mi Trip 2: 12 mi straight south. South component: -12 mi (since it's in the opposite direction of North) Trip 3: 8 mi in a direction 30° north of west. To find the components, we'll use trigonometric functions (sine and cosine) and the given angle. West component: 8 mi * cos(30°) = 6.93 mi (rounded to two decimal places) North component: 8 mi * sin(30°) = 4 mi Now we have the components for each trip: Trip 1: East=10 mi, North=10 mi Trip 2: South=-12 mi Trip 3: West=-6.93 mi, North=4 mi

Calculate the total displacement components

Now we add up the components to find the total East-West and North-South components of the displacement. Total East component: 10 mi (from Trip 1) Total West component: -6.93 mi (from Trip 3) Total North component: 10 mi (from Trip 1) + 4 mi (from Trip 3) = 14 mi Total South component: -12 mi (from Trip 2) Then, combine the East and West components, as well as the North and South components. East-West component: 10 mi - 6.93 mi = 3.07 mi North-South component: 14 mi - 12 mi = 2 mi So, the man's total displacement has a horizontal component of 3.07 mi and a vertical component of 2 mi.

Calculate the magnitude of the resultant displacement

We can find the magnitude of his total displacement by using the Pythagorean theorem on the horizontal and vertical components. Magnitude = sqrt(East-West component^2 + North-South component^2) Magnitude = sqrt(3.07 mi^2 + 2 mi^2) Magnitude = 3.75 mi (rounded to two decimal places) So, the magnitude of the man's total displacement is approximately 3.75 mi.

Calculate the direction of the resultant displacement

For the direction of the displacement, let's find the angle between the displacement vector and the eastward direction. We can do this using the arctangent function (atan) on the ratio of the North-South component to the East-West component. Angle = atan(North-South component / East-West component) Angle = atan(2 mi / 3.07 mi) Angle = 33.1° (rounded to one decimal place) Since the man's total displacement is northeast, the angle with respect to the eastward direction is 33.1°. #Conclusion#The man's resultant displacement has a magnitude of 3.75 mi and is at an angle of 33.1° from the eastward direction (to the northeast).

Key concepts

Displacement Components

In physics, displacement refers to the change in position of an object and is a vector quantity, meaning it has both magnitude and direction. To understand displacement thoroughly, it's important to break it down into its components.
When dealing with any movement that isn't confined to a single line, it's helpful to think in terms of directions within a coordinate plane—most commonly, the East-West and North-South axes. When a movement occurs in a diagonal manner like northeast, each movement can be resolved into its component parts—one vector along the East-West axis and another along the North-South axis.

• For movement towards the East, the component is positive along the East direction.
• Movement towards the North is positive along the vertical North axis.
• Conversely, movement to the West or South is considered negative along their respective axes.

In our example, by resolving the man's trips into East-West and North-South components, we simplify the problem, allowing us to determine the total displacement effectively.

Trigonometry in Physics

Trigonometry provides powerful tools for solving physics problems involving angles and distances. Angles play an essential role in defining directions for vector components.
Physics frequently uses trigonometric functions such as sine, cosine, and tangent to find vector components. When a direction is not perfectly aligned with the North-South or East-West axes, trigonometry helps to determine the respective components.

• **Sine (sin)** helps find the opposite side of a right-angled triangle, essential for determining the vertical component when given an angle.
• **Cosine (cos)** determines the adjacent side, crucial for finding the horizontal component.

For instance, when the direction is 30° north of west:
- The cosine of 30° provides the west component, indicating how much of the distance is in the westward direction.
- The sine of 30° helps find how much the movement stretches north. Using trigonometry not only simplifies problems but also improves understanding of how different directional components contribute to a vector.

Vector Addition

Vector addition allows combining two or more vectors to find a resultant vector. This process is essential in physics to find the net effect of several separate movements.
When vectors are not aligned with each other, adding them directly isn't straightforward. Instead, we should:

• Break each vector down into its components, as discussed earlier.
• Add corresponding components separately (i.e., all East-West components together and all North-South components together).

In the example problem:

• The man’s journey involves multiple legs, each contributing differently along the horizontal and vertical axes.
• By adding the east-west components and the north-south components individually, we arrive at a total resultant displacement.

Calculating the displacement's magnitude and direction often involves using the Pythagorean theorem and arctangent functions to provide complete descriptions of the movement in terms of both length and angle.
Vector addition hence transforms complex navigation problems into manageable calculations.