Question

You are lost at night in a large, open field. Your GPS tells you that you are 122.0 m from your truck, in a direction 58.0${}^{\circ }$ east of south. You walk 72.0 m due west along a ditch. How much farther, and in what direction, must you walk to reach your truck?

Short answer

You need to walk 71.4 m at 63.6° south of west to reach the truck.

Step-by-step solution

Understand the Problem

We need to calculate the additional distance and direction required to reach the truck after walking 72.0 m due west, from our original point 122.0 m away in a direction 58.0° east of south.

Break Down Initial Position

At the starting point, the truck is 122.0 m away at 58.0° east of south. We can break this into components:- Use $$x_0=122.0\sin ({58.0}^{\circ })$$ for the east-west direction.- Use $$y_0=-122.0\cos ({58.0}^{\circ })$$ for the north-south direction since it is south.

Calculate Initial Components

Calculate the components:- East component: $$x_0=122.0\sin ({58.0}^{\circ })=103.6\text{m}$$- South component: $$y_0=-122.0\cos ({58.0}^{\circ })=-64.6\text{m}$$

Apply Displacement

You walk 72.0 m due west. This changes only the east-west component:$$x=103.6-72.0=31.6\text{m}$$The north-south component remains unchanged:$$y=-64.6\text{m}$$

Calculate Final Distance

Use the Pythagorean theorem to find the final distance to the truck from the current position:$$d=\sqrt{(31.6{)}^2+(-64.6{)}^2}=71.4\text{m}$$

Find Direction to Truck

Calculate the angle from the west-east axis using:$$\theta ={\tan }^{-1}\left(\frac{64.6}{31.6}\right)={63.6}^{\circ }$$Since the direction is south of west, the final direction is 63.6° south of west.

Key concepts

Vector Components

Vector components play a crucial role in solving problems involving direction and distance in physics. Imagine vectors as arrows pointing in a certain direction, with both magnitude and direction supplying detailed movement information.
To simplify calculations, these vectors can be broken down into two perpendicular lines or components: east-west and north-south.

• The east-west vector component can often be determined using trigonometry, particularly using the sine function for angles given in a clockwise rotation or from a specific direction.
• Similarly, the north-south vector component might use the cosine function, adjusting the component direction based on whether it involves a southward or northward movement.

In the original problem, calculating these vector components requires understanding how a vector in a specific direction—58° east of south—can be split. For the initial vector from your position toward the truck, apply:

• East-West Component: Use $$x_0=122.0\sin ({58.0}^{\circ })=103.6\text{m}$$
• North-South Component: Use $$y_0=-122.0\cos ({58.0}^{\circ })=-64.6\text{m}$$ contain important directional information. They allow us to determine precise positions on an otherwise flat plane, offering a substantial breakthrough in vector analyses for students.

Trigonometry in Physics

Trigonometry, a branch of mathematics dealing with triangles, shines in physics, especially in vector analysis. By leveraging trigonometry, you can unravel angles and distances with precision.
Physics problems frequently involve right-angled triangles, with an angle and hypotenuse given; trigonometry helps you determine the unknown adjacent or opposite sides. When applying trigonometry in this exercise, consider the following:

• Sine: Use sine ( $\sin$) whenever you need to find the length of the side opposite an angle in a right triangle when the hypotenuse is known.
• Cosine: Incorporate cosine ( $\cos$) for finding the adjacent side in a right-angled triangle.
• Tangent: Use tangent ( $\tan$) to find angles when both the opposite and adjacent side lengths are known.

To calculate a new direction or resolve a force, the process involves identifying which trigonometric function relates to the known and required values. In our problem, you break down the displacement into compass direction components. The understanding of these functions turns trigonometry into an essential tool in physics to make sense of directional data efficiently.

Direction and Displacement Calculation

Direction and displacement form the core of vector analysis in physics problems. Understanding how to calculate final displacement from an initial point using vector components is paramount for unraveling the final position.
Displacement refers not just to how far you've traveled, but also in what direction.
After calculating vector components, apply the Pythagorean theorem:

• To find the total displacement ( $d$), use the formula: $$d=\sqrt{(x^2)+(y^2)}$$. This equation provides the direct path or shortest distance between your starting location and your destination, represented by the current position.

To determine the overall direction relative to a specified axis (often using compass directions):

• Calculate the angle using the tangent function, applying: $$\theta ={\tan }^{-1}\left(\frac{\text{opposite side}}{\text{adjacent side}}\right)$$. This equation helps you find the angular direction in practical terms, such as "south of west." For your position related to the truck, it turns into an accurate directional reading—63.6° south of west in this scenario.

Mastering these calculations can offer profound insights into navigation and vector control across multiple physics applications.