Problem 283 - Elementary Algebra

Question

In the following exercises, solve. Lilah is moving from Portland to Seattle. It takes her three hours to go by train. Mason leaves the train station in Portland and drives to the train station in Seattle with all Lilah's boxes in his car. It takes him 2.4 hours to get to Seattle, driving at 15 miles per hour faster than the speed of the train. Find Mason's speed and the speed of the train.

Step-by-Step Solution

Answer

The train's speed is 60 mph; Mason's speed is 75 mph.

Step 1: Set up the variables

Let the speed of the train be denoted as \( v_t \) (in miles per hour). Mason's speed is 15 miles per hour faster, so let Mason's speed be denoted as \( v_m = v_t + 15 \).

Step 2: Write the distance equation for the train

Using the formula for distance, which is \( \text{distance} = \text{speed} \times \text{time} \), the distance the train travels is: \[ d = v_t \times 3 \text{ hours} \]

Step 3: Write the distance equation for Mason

Similarly, the distance Mason travels in 2.4 hours is: \[ d = v_m \times 2.4 \text{ hours} \]. Substituting \( v_m = v_t + 15 \) gives: \[ d = (v_t + 15) \times 2.4 \]

Step 4: Set the distances equal

Since both the train and Mason travel the same distance, set the two equations equal to each other: \[ v_t \times 3 = (v_t + 15) \times 2.4 \]

Step 5: Solve for the train's speed

Expand and simplify the equation to find \( v_t \):\[ 3v_t = 2.4v_t + 36 \]. Subtract \( 2.4v_t \) from both sides: \[ 0.6v_t = 36 \]. Then divide by 0.6: \[ v_t = 60 \]. So, the train's speed is 60 miles per hour.

Step 6: Find Mason's speed

Now use the train's speed to find Mason's speed: \[ v_m = v_t + 15 = 60 + 15 = 75 \]. So, Mason's speed is 75 miles per hour.

Key Concepts

Algebraic Equations

Algebraic equations are essential for solving many real-world problems, like the example of Lilah and Mason traveling to Seattle. By setting up an equation, we can use known values and relationships to find unknown variables. In our exercise, we defined the speed of the train as a variable, making it easier to solve the problem systematically.
First, we recognized that we have two speeding components: the train and Mason. We turned these into algebraic equations based on their speeds and travel times, which allowed us to set up a relatable and solvable problem structure.

Solving for Variables

Finding the values of variables is a crucial step in algebra. Here, we had to determine the speed of the train and then use that to find Mason's speed.
We started by defining the variables: let the train's speed be denoted as \( v_t \). Mason's speed is 15 miles per hour faster, so it’s \( v_m = v_t + 15 \).
We formed an equation for the distances traveled by both Mason and the train and set them equal because they both traveled the same distance. Solving these equations involved algebraic manipulation to isolate the variable \( v_t \), allowing us to find the train's speed first and then using it to compute Mason’s speed efficiently.

Real-Life Math Applications

Many everyday situations require the application of math principles. The travel problem involving Lilah and Mason shows how these concepts can help in real-life scenarios. Whether it's calculating travel times, distances, or speeds, these math skills are highly practical.
Understanding how to set up equations based on real-world parameters ensures that we can solve problems efficiently, making sense of complexities like travel logistics. This exercise also demonstrates how having solid mathematical skills can make planning and decision-making much simpler and more logical.

Rate-Time-Distance Relationships

The relationship between rate, time, and distance is central to many travel-related problems. The formula we used, \( \text{distance} = \text{speed} \times \text{time} \), is fundamental in physics and everyday calculations. Understanding how to manipulate this formula allows us to solve for any unknown variable when the other two are known.
In the context of our exercise, we used:

• the travel time of the train (3 hours) and Mason (2.4 hours)
• the unknown speeds of both
• and the consistent distance they both traveled.

This approach shows how closely rate, time, and distance are interlinked, reinforcing the importance of mastering these concepts to solve various practical problems effectively.