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As a test of orienteering skills, your physics class holds a contest in a large, open field. Each contestant is told to travel 20.8 m due north from the starting point, then 38.0 m due east, and finally 18.0 m in the direction 33.0\(^{\circ}\) west of south. After the specified displacements, a contestant will find a silver dollar hidden under a rock. The winner is the person who takes the shortest time to reach the location of the silver dollar. Remembering what you learned in class, you run on a straight line from the starting point to the hidden coin. How far and in what direction do you run?

Short Answer

Expert verified
You should run approximately 57.1 m at an angle of 12.0° south of east.

Step by step solution

01

Break Down the Movements

First, let's determine each movement separately. Your initial movement is 20.8 m due north, followed by 38.0 m due east. Finally, the movement is 18.0 m at an angle of 33.0° west of south.
02

Convert Directions to Vector Components

We convert each displacement into vector components. The north movement is purely in the y-direction: \[\begin{align*}R_1 &= (0, 20.8)\end{align*}\]The east movement, purely in the x-direction: \[\begin{align*}R_2 &= (38.0, 0)\end{align*}\]For the movement 33.0° west of south, we first adjust to refer to standard position: this angle is equivalent to 33.0° west of the vertical south axis.
03

Calculate the Components for the Third Movement

The third vector needs conversion using trigonometric functions. Since the movement is westward of the south axis, the x-component will be negative (west) and the y-component negative (south):\[\begin{align*}R_3^x &= -18.0 \cdot \sin(33.0°) \R_3^y &= -18.0 \cdot \cos(33.0°)\end{align*}\]
04

Sum the Components of All Moves

The total displacement vector is the sum of all individual vector components:\[\begin{align*}R_x &= 0 + 38.0 + (-18.0 \cdot \sin(33.0°)) \R_y &= 20.8 + 0 + (-18.0 \cdot \cos(33.0°))\end{align*}\]Calculate these components to get numerical values.
05

Calculate the Magnitude of the Total Displacement

The magnitude \( R \) of your straight-line path is calculated using the Pythagorean theorem:\[R = \sqrt{(R_x)^2 + (R_y)^2}\] Plug in the values calculated in Step 4 to find the distance directly to the hidden dollar.
06

Calculate the Direction of the Total Displacement

The direction \( \theta \) of your path is found using the arctangent function:\[\theta = \arctan\left(\frac{R_y}{R_x}\right)\]determine whether the angle should be adjusted based on which quadrant it lies in to report a bearing or angle from the starting point.

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Key Concepts

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

Displacement
Displacement in physics refers to the change in position of an object. Imagine it as a straight line connecting the starting and ending points of an object's journey, irrespective of how complex the actual path traveled might be. In the orienteering exercise described, displacement is the direct line from the starting point to the location of the silver dollar.
Displacement is typically represented as a vector, meaning it has both magnitude (how far) and direction (which way).
  • Magnitude: It tells us the straight-line distance from start to finish, often calculated using the Pythagorean theorem when there are right-angle components involved.
  • Direction: This is the angle or path that leads directly from the starting point to the end point, often determined using trigonometric functions.
In this exercise, finding the displacement means identifying the shortest and most efficient run directly to the hidden coin by considering the combined effect of all the moves.
Trigonometric Functions
Trigonometric functions are essential in understanding movement and direction, especially when dealing with angles. These functions help break down any movement into components that can be easily analyzed.
The key trigonometric functions used in displacement calculations are sine, cosine, and tangent.
  • Sine ( \( \sin \) ): Used to find the opposite side of an angle in a right triangle. In our exercise, it aids in determining the horizontal component of the third vector.
  • Cosine ( \( \cos \) ): Used to find the adjacent side of an angle in a right triangle. It helps figure out the vertical component of the third move in this scenario.
  • Tangent ( \( \tan \) ): Defined as the ratio of sine to cosine, this function is instrumental in calculating the angle of displacement, hence determining the direction.
By applying these functions, you can convert any angled movement into easily understood x and y component vectors, simplifying the whole process of calculating displacement.
Orienteering
Orienteering involves navigating from one point to another using your map reading, compass skills, and decision-making abilities. It's somewhat like a treasure hunt, where the treasure is a successful and efficient route to a destination.
In this exercise, the physics class contest mimics orienteering through a preset series of movements across a field.
  • Map Reading: Understanding the given problem as a map of movements. Moving north, east, and finally at an angle requires mental mapping and converting these moves into a single resultant vector.
  • Using a Compass: Functions as your "trigonometric calculator," guiding through angles and directions while converting them into practical components.
  • Efficient Route Finding: By solving for the displacement using vector addition, one calculates a shortcut, directly reaching the hidden coin.
Orienteering blends well with vector math in physics, practicing calculation and navigation skills, both critical in real-world journeys!

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Most popular questions from this chapter

The most powerful engine available for the classic 1963 Chevrolet Corvette Sting Ray developed 360 horsepower and had a displacement of 327 cubic inches. Express this displacement in liters (L) by using only the conversions 1 L = 1000 cm\(^3\) and 1 in. = 2.54 cm.

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A fence post is 52.0 m from where you are standing, in a direction 37.0\(^{\circ}\) north of east. A second fence post is due south from you. How far are you from the second post if the distance between the two posts is 68.0 m?

Find the angle between each of these pairs of vectors: (a) \(\overrightarrow{A}\) = \(-\)2.00\(\hat{\imath}\) \(+\) 6.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 2.00\(\hat{\imath}\) \(-\) 3.00\(\hat{\jmath}\) (b) \(\overrightarrow{A}\) = 3.00\(\hat{\imath}\) \(+\) 5.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 10.00\(\hat{\imath}\) \(+\) 6.00\(\hat{\jmath}\) (c) \(\overrightarrow{A}\) = \(-\)4.00\(\hat{\imath}\) \(+\) 2.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 7.00\(\hat{\imath}\) \(+\) 14.00\(\hat{\jmath}\)

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