Question

Two cars traveled the same distance. One car traveled at 50 \(\mathrm{mph}\) and the other car traveled at \(60 \mathrm{mph}\). It took the slower car 50 minutes longer to make the trip. How long did it take the faster car to make the trip?

Short answer

It took the faster car \(4\) hours and \(10\) minutes to make the trip.

Step-by-step solution

Assign Variables

Let's assign variables to the unknowns in the problem: Let \(t\) be the time it took the faster car to make the trip (in hours). Let \(t + \frac{50}{60}\) be the time it took the slower car to make the trip (in hours). We add \(\frac{50}{60}\) since the slower car took 50 minutes longer, and we want the time in hours.

Set Up the Equation

Since the two cars traveled the same distance, we can equate their distances using the formula: distance = speed × time. For the slower car, distance = 50 mph × (t + 50/60) hours For the faster car, distance = 60 mph × t hours Now, we can set up the equation: 50 × (t + 50/60) = 60 × t

Solve the Equation

To solve for \(t\), we can first simplify the equation: 50 × (t + 5/6) = 60 × t Now we can distribute the 50: 50t + 250/6 = 60t Now, subtract 50t from both sides of the equation: 250/6 = 10t Next, divide both sides by 10: 25/6 = t

Convert Time to Hours and Minutes

Now that we have the time in hours, we can convert it to hours and minutes: \(t = \frac{25}{6} = 4 \frac{1}{6}\) hours So, it took the faster car 4 hours and 10 minutes to make the trip.

Key concepts

Distance-Speed-Time

Understanding the relationship between distance, speed, and time is crucial in many real-world applications. The formula connecting these three important elements is:

• **Distance = Speed × Time**

This straightforward equation helps us determine any one of the variables if the others are known.
In the exercise, two cars travel the same distance but at different speeds. By using their known speeds, 50 mph and 60 mph, and the fact that they cover the same distance, we can set up an equation to figure out the time taken.
Applying this formula to both cars, we can express distance as:

• **Slower Car**: \(50 ext{ mph} \times (t + \frac{50}{60}) ext{ hours}\)
• **Faster Car**: \(60 ext{ mph} \times t ext{ hours}\)

Understanding how these pieces fit together allows us to solve for unknowns, like how long it took for the faster car to complete its journey.

Equation Solving

Once you have an equation representing a problem, the next step is to solve it. In this case, the equation derived from our distance-speed-time knowledge is: \[ 50 \times (t + \frac{50}{60}) = 60 \times t \]To solve it, follow these steps:

• Simplify the equation by distributing: \(50t + \frac{250}{6} = 60t\)
• Then, isolate the variable by subtracting \(50t\) from both sides: \(\frac{250}{6} = 10t\)
• Finally, solve for \(t\) by dividing both sides by 10: \(t = \frac{25}{6}\)

Solving equations involves simplifying expressions, applying arithmetic operations, and isolating the variable of interest. This process is a fundamental building block of algebra, which is itself a foundation for both calculus and pre-calculus problems.

Unit Conversion

Understanding unit conversion is key when working with problems involving different units of measure. In the exercise, time is given in minutes and needs to be converted to hours to be compatible with the speed provided in mph (miles per hour).
Here’s how you can convert:

• 1 hour = 60 minutes
• To convert 50 minutes to hours, use the fraction \(\frac{50}{60}\) hours.

Converting units ensures consistency and accuracy in calculations, especially when combining different measurements. Using this concept, we accurately adjust the problem to a single unit system, making it easier to solve for the desired variable.

Variable Assignment

In any problem, it's essential to clearly define and assign variables. This is like setting the stage for the solution process. For the original exercise, we need to represent time with variables to develop an equation.
Here's how you can assign variables effectively:

• Let \(t\) represent the time it takes for the faster car (in hours).
• For the slower car, add the extra 50 minutes it takes, which converts to \(t + \frac{50}{60}\) hours.

Assigning variables helps to transform word problems into mathematical equations, which are often much simpler to solve. Defining variables precisely is crucial for establishing a clear path from the problem statement to an eventual solution.