Question

A person desires to reach a point that is \(3.42 \mathrm{~km}\) from her present location and in a direction that is \(35.0^{\circ}\) north of east. However, she must travel along streets that go either northsouth or east-west. What is the minimum distance she could travel to reach her destination?

Short answer

To find the minimum distance to be traveled, add up the calculated horizontal (east) and vertical (north) distances.

Step-by-step solution

Break destination into components

The destination is at \(35.0^{\circ}\) north of east, which means we can break it into east and north components. The north component denotes how much she will travel in the north direction, while the east component denotes east direction travel. The east component can be calculated using cosine of the angle, which gives the adjacent side in a right triangle. Hence, east component = \(3.42 \mathrm{~km} \times \cos(35.0^{\circ})\). Similarly, the north component can be calculated using sine of the angle, which gives the opposite side in a right triangle. Hence, north component = \(3.42 \mathrm{~km} \times \sin(35.0^{\circ})\).

Calculate individual distances

Now, calculate the distances based on the above formulae. The east component corresponds to the horizontal distance, and the north component corresponds to the vertical distance. Use a calculator to fetch these values.

Calculate total distance travelled

Once we get the individual distances traveled in the east and north direction, add them together. This gives the total distance traveled by the person in the city grid.

Interpretation of the result

The sum of the distances traveled in the east and north directions will give the minimum distance required to reach the desired spot by traveling either north-south or east-west.

Key concepts

Trigonometry in Physics

Trigonometry is a mathematical tool used to relate angles to side lengths in right-angled triangles. In physics, trigonometry helps to break vectors into components which specially aids in calculating motion and direction.

In the given exercise, the person needs to travel to a location at an angle () 35.0^{\circ}) north of east. To simplify this so that she can move only along streets running north-south or east-west, we break this vector into its components: the east component and the north component.

A vector directed at an angle can be split using the trigonometric functions cosine and sine. The east component (adjacent side) of the vector is found by using the cosine function:\[ \text{East Component} = 3.42 \, \text{km} \times \cos(35.0^{\circ}) \]
Similarly, the north component (opposite side) of the vector is found by using the sine function:\[ \text{North Component} = 3.42 \, \text{km} \times \sin(35.0^{\circ}) \]
These trigonometric components simplify navigating practical paths like city streets much more straightforward.

Distance Calculation

Distance calculation on a grid requires clear breakdowns. After determining the vector components using trigonometry, calculating the respective distances is straightforward.

The person will travel individually the northwest and northwest distances obtained from the components. This exercise does not deal with diagonal travel but rather focuses on moving along orthogonal directions.

Once you have the vector components for east and north, use them to calculate the minimum travel distance. Simply take the sum of the east and north components, as each represents direct travel along those directions on the grid:

• The east component is found with \( 3.42 \, \text{km} \times \cos(35.0^{\circ}) \).
• The north component uses \( 3.42 \, \text{km} \times \sin(35.0^{\circ}) \).

Adding these values will ensure you measure the minimal distance along the designated paths in the city grid.

City Grid Navigation

The concept of city grid navigation comes into play when movement is restricted to perpendicular directions — often the case in cities organized in a grid pattern. This gives rise to a unique challenge, requiring navigation using predetermined pathways.

Imagine driving in a city with strict grid-like streets, where you can only make right-angle turns. The goal is to move efficiently from point A to point B using the shortest path possible, without diagonal shortcuts.

The exercise illustrates using vector components to model such a journey. Rather than flying directly at and are dissecting the travel into only north-south and east-west paths. This way, each movement follows the constraints of a typical city layout, making the distance-oriented.

Using the trigonometric approach to break down vectors is especially useful for navigation problems. It prepares you for real-world scenarios like delivery routes or traveling large cityscapes, demonstrating precise and efficient path planning.