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Train Travel. A train leaves Union Station for Central Station, \(216 \mathrm{km}\) away, at 9 A.M. One hour later, a train leaves Central Station for Union Station. They meet at noon. If the second train had started at \(9 \mathrm{A.M.}\) and the first train at \(10: 30 \mathrm{A.M.}\) they would still have met at noon. Find the speed of each train.

Short Answer

Expert verified
The speed of the first train is 36 km/hr, and the speed of the second train is 54 km/hr.

Step by step solution

01

- Define Variables

Let the speeds of the two trains be represented by the variables: \(v_1\) for the first train, and \(v_2\) for the second train.
02

- Set Up the Time for Meeting

The first train travels for 3 hours (from 9 A.M. to noon), and the second train travels for 2 hours (from 10 A.M. to noon). Use the information given to find the distance each train has traveled when they meet: \(d_1 = 3v_1\) for the first train, and \(d_2 = 2v_2\) for the second train.
03

- Create an Equation for Distance

The sum of the distances covered by both trains when they meet should add up to 216 km. Therefore, the equation is: \[3v_1 + 2v_2 = 216\]
04

- Utilize the Alternate Scenario

According to the alternate scenario, if the second train departed at 9 A.M. and the first train at 10:30 A.M., they would meet at noon. Therefore, in this case, the second train travels 3 hours while the first one travels 1.5 hours. Form the equation for this situation:\[3v_2 + 1.5v_1 = 216\]
05

- Solve the System of Equations

We have two equations now: \[3v_1 + 2v_2 = 216\]\[3v_2 + 1.5v_1 = 216\] To solve for the speeds of the trains, we'll use substitution or elimination method. Here we'll use the elimination method.
06

- Eliminate One Variable

To eliminate one variable, multiply the second equation by 2 to make the coefficients of \(v_1\) equal:\[6v_2 + 3v_1 = 432\] Subtract the first equation from this new equation: \[6v_2 + 3v_1 - (3v_1 + 2v_2) = 432 - 216\] Simplifying gives: \[4v_2 = 216\]
07

- Solve for \(v_2\)

Divide both sides of the equation by 4: \[v_2 = 54\] So the speed of the second train is 54 km/hr.
08

- Solve for \(v_1\)

Substitute \(v_2 = 54\) back into the first equation: \[3v_1 + 2(54) = 216\] \[3v_1 + 108 = 216\] Subtract 108 from both sides: \[3v_1 = 108\] Divide by 3: \[v_1 = 36\] So the speed of the first train is 36 km/hr.

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Key Concepts

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

Intermediate Algebra
Intermediate algebra is a fundamental concept that helps us solve problems involving unknown values. In our train travel problem, we use variables to represent the speeds of the two trains. This allows us to form equations based on the given conditions. By using techniques like substitution or elimination, we can solve these equations to find the values of the unknown variables.

In this exercise, variables such as \(v_1\) and \(v_2\) are used for the speeds of the trains. Then, we create equations to represent the distances traveled by each train based on their speed and travel time. Intermediate algebra helps us break down the information and systematically solve the problem.
System of Equations
A system of equations is a set of two or more equations that have common variables. Solving systems of equations involves finding the values of the variables that satisfy all the equations simultaneously. This method is crucial in problems like train travel, where multiple conditions must be met at the same time.

In our case, we have two equations:
\[3v_1 + 2v_2 = 216\] \[3v_2 + 1.5v_1 = 216\]
These equations arise from considering two different travel scenarios where distances add up to the same value of 216 km. To solve this system, we can use methods such as the substitution method or the elimination method. Here, the elimination method was employed to eliminate one variable and solve for the other. Once one variable was found, it was substituted back into one of the original equations to find the other variable.
Distance and Speed Calculations
Distance and speed calculations are essential in understanding the relationship between how fast something is moving and how far it will travel over a given period. In problems involving moving objects like trains, it is common to use the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \]

For example, in our train travel problem:
- The first train travels for 3 hours at a speed of \(v_1\), covering a distance of \(d_1 = 3v_1\).
- The second train travels for 2 hours at a speed of \(v_2\), covering a distance of \(d_2 = 2v_2\).

When the two trains meet, the sum of the distances covered by both trains is equal to the total distance between the stations, which is 216 km. By setting up equations based on these relationships, we can solve for the speeds of the trains.

Understanding how to break down the problem and apply the distance formula is key to solving similar problems. This involves finding the time each train has traveled, multiplying it by their respective speeds, and ensuring the total distance matches the given problem condition.

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