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Chapter 5: Problem 62

Two particles begin at the same point and move at different speeds along a circular path of circumference \(280 \mathrm{ft}\). Moving in opposite directions, they pass in \(10 \mathrm{sec} .\) Moving in the same direction, they pass in \(70 \mathrm{sec} .\) Find the speed of each particle.

Short Answer

Expert verified
The speeds of the particles are 16 ft/s and 12 ft/s.

Step by step solution

01

Define the problem

Define the speeds of the two particles as \(v_1\) and \(v_2\). Let the circumference of the circular path be \(C = 280 \text{ ft}\). Given the times it takes for the particles to pass each other when moving in opposite and in the same direction, let's denote these times as \(t_{\text{opposite}} = 10 \text{s}\) and \(t_{\text{same}} = 70 \text{s}\).
02

Set up equations for opposite directions

When the particles move in opposite directions, the relative speed is \(v_1 + v_2\). They pass each other every \(10 \text{s}\), so:\[ (v_1 + v_2) \times 10 = 280 \text{ ft} \]Thus,\[ v_1 + v_2 = 28 \text{ ft/s} \]
03

Set up equations for the same direction

When the particles move in the same direction, the relative speed is \(|v_1 - v_2|\). They pass each other every \(70 \text{s}\), so:\[ |v_1 - v_2| \times 70 = 280 \text{ ft} \]Thus,\[ |v_1 - v_2| = 4 \text{ ft/s} \]
04

Solve the system of equations

Solve the system of equations obtained from steps 2 and 3:\[ v_1 + v_2 = 28 \text{ ft/s} \]\[ |v_1 - v_2| = 4 \text{ ft/s} \]Since we need to find the positive solutions, consider two cases:- Case 1: \(v_1 - v_2 = 4\)- Case 2: \(v_2 - v_1 = 4\)
05

Address Case 1

For \(v_1 - v_2 = 4\):Adding this to the equation \(v_1 + v_2 = 28\):\[2v_1 = 32\]\[v_1 = 16 \text{ ft/s}\]Substituting \(v_1\) back into \(v_1 + v_2 = 28\):\[16 + v_2 = 28\]\[v_2 = 12 \text{ ft/s}\]
06

Address Case 2

For \(v_2 - v_1 = 4\):Adding this to the equation \(v_1 + v_2 = 28\):\[2v_2 = 32\]\[v_2 = 16 \text{ ft/s}\]Substituting \(v_2\) back into \(v_1 + v_2 = 28\):\[v_1 + 16 = 28\]\[v_1 = 12 \text{ ft/s}\]
07

Conclude the solution

From both cases, we get that the speed of the particles is \(v_1 = 16 \text{ ft/s}\) and \(v_2 = 12 \text{ ft/s}\) or \(v_1 = 12 \text{ ft/s}\) and \(v_2 = 16 \text{ ft/s}\).

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Key Concepts

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

relative speed
Relative speed is a concept used to determine how fast two objects are moving with respect to each other. When two objects move in opposite directions, their relative speed is the sum of their individual speeds. On the other hand, if they move in the same direction, their relative speed is the difference between their speeds. In this exercise, we used these principles to set up equations based on the given times.
circular motion
Circular motion occurs when an object moves along the circumference of a circle. Important parameters in circular motion include the circumference, usually denoted as C, and the time taken to complete a certain part of the path. Here, two particles move along a circular path with a circumference of 280 ft. Their interaction times provided key information to determine their speeds.
systems of equations
A system of equations involves multiple equations that are solved together. To find the speeds of our particles, we set up two equations: one for the case when they move in opposite directions and another for when they move in the same direction. The equations were:
  • \( v_1 + v_2 = 28 \text{ ft/s} \)
  • \( |v_1 - v_2| = 4 \text{ ft/s} \)
These equations were then simultaneously solved to find the values of \( v_1 \) and \( v_2 \).
algebraic problem-solving
Algebraic problem-solving involves defining variables, setting up equations, and solving them. The process generally includes:
  • Identifying given information and what needs to be found.
  • Establishing relationships between the variables.
  • Setting up equations based on these relationships.
  • Solving the equations systematically.
In this problem, we finalized the solution by solving a system of linear equations. This step-by-step method is fundamental to tackling algebra problems effectively.

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Most popular questions from this chapter

A patient undergoing a heart scan is given a sample of fluorine- \(18\left({ }^{18} \mathrm{~F}\right)\). After \(4 \mathrm{hr}\), the radioactivity level in the patient is \(44.1 \mathrm{MBq}\) (megabecquerel). After \(5 \mathrm{hr}\), the radioactivity level drops to \(30.2 \mathrm{MBq}\). The radioactivity level \(Q(t)\) can be approximated by \(Q(t)=Q_{0} e^{-k t},\) where \(t\) is the time in hours after the initial dose \(Q_{0}\) is administered. a. Determine the value of \(k\). Round to 4 decimal places. b. Determine the initial dose, \(Q_{0}\). Round to the nearest whole unit. c. Determine the radioactivity level after \(12 \mathrm{hr}\). Round to 1 decimal place.

Solve the system using any method. $$ \begin{array}{l} 5(2 x+y)=y-x-8 \\ x-\frac{3}{2} y=\frac{5}{2} \end{array} $$

Solve the system of equations by using the addition method. (See Examples \(3-4)\) $$ \begin{array}{l} 5 x-2 y=-2 \\ 3 x+4 y=30 \end{array} $$

a. Sketch the lines defined by \(y=2 x\) and \(y=-\frac{1}{2} x+5\) b. Find the area of the triangle bounded by the lines in part (a) and the \(x\) -axis.

Refer to Section 2.5 for a review of linear cost functions and linear revenue functions. A cleaning company charges \(\$ 100\) for each office it cleans. The fixed monthly cost of \(\$ 480\) for the company includes telephone service and the depreciation on cleaning equipment and a van. The variable cost is \(\$ 52\) per office and includes labor, gasoline, and cleaning supplies. a. Write a linear cost function representing the cost \(C(x)\) (in \(\$$ ) to clean \)x\( offices per month. b. Write a linear revenue function representing the revenue \)R(x)\( (in \$) for cleaning \)x$ offices per month. c. Determine the number of offices to be cleaned per month for the company to break even. d. If 28 offices are cleaned, will the company make money or lose money?
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