Question

A truck travels 350 mi to a delivery point, unloads, and, now empty, returns to the starting point at a speed 8.00 milh greater than on the outward trip. What was the speed of the outward trip if the total round-trip driving time was \(14.4 \mathrm{h} ?\)

Short answer

The speed of the outward trip was 40 milh.

Step-by-step solution

Define Variables

Let's denote the speed of the truck on the outward trip as 's' milh. Since the truck travels 8 milh faster on the return trip, the speed for the return trip is 's + 8' milh.

Write Time Expressions

The time to cover a distance is given by the formula 'time = distance/speed'. So, for the outward trip, the time is 350/s hours, and for the return trip, the time is 350/(s+8) hours.

Set Up the Equation using Total Time

The total driving time is the sum of the times for both trips, which equals 14.4 hours. Therefore, the equation is 350/s + 350/(s + 8) = 14.4.

Clear the Denominators

To solve the equation, find a common denominator and clear the fractions. The common denominator is s(s + 8), so we multiply each term by this to get 350(s + 8) + 350s = 14.4s(s + 8).

Expand and Simplify

Distribute and combine like terms to simplify the equation: 350s + 2800 + 350s = 14.4s^2 + 115.2s. This simplifies to 700s + 2800 = 14.4s^2 + 115.2s.

Rearrange into a Quadratic Equation

Subtract 700s and 2800 from both sides to get 0 = 14.4s^2 + 115.2s - 700s - 2800. This simplifies to 0 = 14.4s^2 - 584.8s - 2800.

Use the Quadratic Formula to Solve for 's'

Apply the quadratic formula, s = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = 14.4, b = -584.8, and c = -2800. This will give us the possible values for 's'.

Solve for 's'

After calculation, you should get two potential solutions for 's'. Only one will be a reasonable speed for a truck (a positive value less than the total distance divided by the total time).

Key concepts

Quadratic Equations

Quadratic equations are a fundamental component of algebra and appear in a variety of real-world situations, including round-trip motion problems. A quadratic equation is characterized by the standard form \( ax^2+bx+c=0 \), where \( a \), \( b \), and \( c \), are coefficients and \( x \) represents the variable or unknown that we aim to solve for.

In the context of our truck problem, the quadratic equation comes from relating the distances and speeds of the truck on its outward and return journeys, encapsulating these into an algebraic expression that matches the total driving time. By rearranging and manipulating this expression, we inevitably encounter a quadratic equation that needs to be solved.

Solving quadratic equations can be achieved through various methods, such as factoring, completing the square, or using the quadratic formula. The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is particularly useful in problems that are difficult to factor and guarantees a solution for any quadratic equation, provided \( b^2-4ac \) is non-negative.

It's essential to evaluate both potential solutions from the quadratic formula in these motion problems, as sometimes one answer may not make physical sense (such as a negative speed), leaving the other as the reasonable solution.

Problem-solving in Physics

Physics problems often involve motion and can be analyzed using algebraic expressions and equations to describe relationships between variables such as speed, distance, and time. The truck problem presented is a classic example of uniform motion, requiring the application of the fundamental distance formula \( \text{distance} = \text{speed} \times \text{time} \).

In this scenario, motion is restricted to a round-trip, introducing constraints that aid in setting up the equations. By breaking down the problem into smaller parts - the outward trip and the return trip - and establishing relationships between speed, time, and distance for each leg, we create a system of equations that encapsulates the entire journey's dynamics.

Understanding Variables

It is crucial to clearly define your variables at the beginning and maintain consistency throughout the problem-solving process. With everything defined, we translate the physical situation into mathematical terms, usually ending up with an algebraic or a quadratic equation that needs to be solved. The solution to this equation is interpreted within the context of the problem to ensure it aligns with physical reality.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, variables, and operators (like addition and subtraction). They are essential tools for translating real-world problems into solvable equations. In problems like our truck example, setting up the correct algebraic expressions is the first step towards finding the solution.

Creating the expressions requires a clear understanding of the relationships between different quantities. For the truck problem, we first express the time of each part of the trip as a ratio of distance to speed. These expressions are then used in combination, reflective of the total travel time, establishing an equation that must be satisfied.

Working with Fractions

In dealing with round-trip motion problems, you will almost always encounter fractions. Having a strategy for handling them, such as finding a common denominator to combine terms or to clear fractions altogether, is vital for simplifying the algebraic expressions into a form that can be easily managed and solved, typically leading to a quadratic equation when dealing with constant speeds and fixed distances.