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If you walk \(35 \mathrm{~km}\) at an angle \(25^{\circ}\) counterclockwise from east, and then \(22 \mathrm{~km}\) at \(230^{\circ}\) counterclockwise from east, find the distance and direction from your starting point to your destination. (answer check available at lightandmatter.com)

Short Answer

Expert verified
The distance is approximately 17.71 km, and the direction is about 353° counterclockwise from east.

Step by step solution

01

Determine the Cartesian Components

To find the cartesian components for the movements, we use trigonometric functions. First, for the 35 km walk at a 25-degree angle:\[x_1 = 35 \cos(25^{\circ}), \quad y_1 = 35 \sin(25^{\circ})\]Second, for the 22 km walk at a 230-degree angle:\[x_2 = 22 \cos(230^{\circ}), \quad y_2 = 22 \sin(230^{\circ})\]
02

Calculate the x-components

Using the formulas from Step 1, calculate the x-components:\[x_1 = 35 \cos(25^{\circ}) \approx 31.71\]\[x_2 = 22 \cos(230^{\circ}) \approx -14.14\]The total x-component is the sum of the two: \[x_{ ext{total}} = x_1 + x_2 \approx 31.71 - 14.14 = 17.57\]
03

Calculate the y-components

Using the formulas from Step 1, calculate the y-components:\[y_1 = 35 \sin(25^{\circ}) \approx 14.79\]\[y_2 = 22 \sin(230^{\circ}) \approx -16.96\]The total y-component is the sum of the two:\[y_{ ext{total}} = y_1 + y_2 \approx 14.79 - 16.96 = -2.17\]
04

Find the Resultant Displacement

Using the total components, calculate the magnitude of the resultant displacement:\[R = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} \approx \sqrt{17.57^2 + (-2.17)^2} \approx 17.71\]
05

Determine the Direction

To find the direction of the resultant displacement relative to the east, use the inverse tangent function:\[\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) \approx \tan^{-1}\left(\frac{-2.17}{17.57}\right) \approx -7.0^{\circ}\]Since the angle is negative and in the fourth quadrant, the angle with respect to east going counterclockwise is:\[360^{\circ} - 7.0^{\circ} = 353^{\circ}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometry
Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is an essential tool in vector addition, which is why we often use trigonometric functions like sine and cosine when breaking down vectors into Cartesian components.

In the given exercise, you walked at certain angles from a specified direction. To determine how far you've walked in each horizontal (x) and vertical (y) direction, we need trigonometry. When you walk 35 km at a 25-degree angle, you can split this movement into two parts: one along the east-west direction and the other along the north-south direction.
The cosine function helps determine the east-west component, while the sine function helps with the north-south component.
  • The equation for the x-component using cosine is: \(x = ext{distance} \times \cos( ext{angle})\)
  • The equation for the y-component using sine is: \(y = ext{distance} \times \sin( ext{angle})\)
We used these trigonometric functions for both the 35 km and 22 km walks, thus allowing us to later calculate the resultant displacement.
Cartesian Components
Cartesian components refer to how we express a vector in terms of its horizontal (x) and vertical (y) components, especially when dealing with vectors on an xy-plane. This is very useful in simplifying calculations involved in vector addition.

In the exercise, each walk was not in a straight line due north or east, so we need to convert them. Using trigonometry, we convert the magnitude of a vector and the angle into x and y components using the wand direction to calculate the components. By breaking each segment of your walk into x and y components:
  • The first walk is split into:
    • \(x_1 = 35 \cos(25^{\circ})\)
    • \(y_1 = 35 \sin(25^{\circ})\)
  • The second walk is:
    • \(x_2 = 22 \cos(230^{\circ})\)
    • \(y_2 = 22 \sin(230^{\circ})\)
Once these components are calculated, adding them together will provide the total x and y components, which can then be converted back into a single resultant vector. This method simplifies finding the overall direction and length of displacement. Cartesian components provide a clearer understanding of complex movement patterns.
Resultant Displacement
Resultant displacement represents the net distance and direction you end up at from your starting point after moving along different paths. It's what you get when you add two or more vectors together. After breaking the vectors into Cartesian components, you can find the total displacement.

The step-by-step solution combines the Cartesian components to find the resultant displacement.
The displacement's magnitude, which is the straight-line directed path from your starting position to your final position, is calculated using the Pythagorean theorem:
  • \(R = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}\)
In this case, it equals approximately 17.71 km. Resultant displacement not only considers how far you've traveled overall but also the exact direction you're facing relative to a standard direction, like east.
Finally, the overall direction of the resultant displacement is found using the inverse tangent function, giving the angle relative to the specified axis, which rounds to 353 degrees. This means you're almost headed directly south-east relative to the starting point. Resultant displacement provides a comprehensive understanding of your overall movement after a series of directional changes.

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Most popular questions from this chapter

Many fish have an organ known as a swim bladder, an air-filled cavity whose main purpose is to control the fish's buoyancy an allow it to keep from rising or sinking without having to use its muscles. In some fish, however, the swim bladder (or a small extension of it) is linked to the ear and serves the additional purpose of amplifying sound waves. For a typical fish having such an anatomy, the bladder has a resonant frequency of \(300 \mathrm{~Hz}\), the bladder's \(Q\) is 3 , and the maximum amplification is about a factor of 100 in energy. Over what range of frequencies would the amplification be at least a factor of \(50 ?\)

Someone tells you she knows of a certain type of Central American earthworm whose skin, when rubbed on polished diamond, has \(\mu_{k}>\mu_{s}\). Why is this not just empirically unlikely but logically suspect?

A ball of mass \(2 m\) collides head-on with an initially stationary ball of mass \(m\). No kinetic energy is transformed into heat or sound. In what direction is the mass- \(2 m\) ball moving after the collision, and how fast is it going compared to its original velocity? \hwans\\{hwans:twotoonecollision\\}

(a) A ball is thrown straight up with velocity \(v\). Find an equation for the height to which it rises.(answer check available at lightandmatter.com) (b) Generalize your equation for a ball thrown at an angle \(\theta\) above horizontal, in which case its initial velocity components are \(v_{x}=v \cos \theta\) and \(v_{y}=v \sin \theta\).(answer check available at lightandmatter.com)

Today's tallest buildings are really not that much taller than the tallest buildings of the \(1940^{\prime}\). One big problem with making an even taller skyscraper is that every elevator needs its own shaft running the whole height of the building. So many elevators are needed to serve the building's thousands of occupants that the elevator shafts start taking up too much of the space within the building. An alternative is to have elevators that can move both horizontally and vertically: with such a design, many elevator cars can share a few shafts, and they don't get in each other's way too much because they can detour around each other. In this design, it becomes impossible to hang the cars from cables, so they would instead have to ride on rails which they grab onto with wheels. Friction would keep them from slipping. The figure shows such a frictional elevator in its vertical travel mode. (The wheels on the bottom are for when it needs to switch to horizontal motion.) (a) If the coefficient of static friction between rubber and steel is \(\mu_{s}\), and the maximum mass of the car plus its passengers is \(M\), how much force must there be pressing each wheel against the rail in order to keep the car from slipping? (Assume the car is not accelerating.)(answer check available at lightandmatter.com) (b) Show that your result has physically reasonable behavior with respect to \(\mu_{s} .\) In other words, if there was less friction, would the wheels need to be pressed more firmly or less firmly? Does your equation behave that way?

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