Problem 59 - Basic Technical Mathematics with Calculus

Question

A hydrofoil made the round-trip of \(120 \mathrm{km}\) between two islands in \(3.5 \mathrm{h}\) of travel time. If the average speed going was \(10 \mathrm{km} / \mathrm{h}\) less than the average speed returning, find these speeds.

Step-by-Step Solution

Answer

Average speed going: 30 km/h; average speed returning: 40 km/h.

Step 1: Define Variables

Let's define the variables: let \( v_1 \) be the average speed going to the island and \( v_2 \) be the average speed returning. According to the problem, the average speed going \( v_1 \) is 10 km/h less than the average speed returning \( v_2 \), so \( v_1 = v_2 - 10 \).

Step 2: Write an Equation Based on Distance

The total distance for the round-trip is 120 km, so the distance each way is 60 km. The travel time for the whole trip is 3.5 hours. The time is given by \( \text{time} = \frac{\text{distance}}{\text{speed}} \). Since \( \text{total time} = \text{time going} + \text{time returning} \), write: \[ \frac{60}{v_1} + \frac{60}{v_2} = 3.5 \]

Step 3: Substitute \( v_1 \) and Simplify

Substitute \( v_1 = v_2 - 10 \) into the equation from Step 2: \[ \frac{60}{v_2 - 10} + \frac{60}{v_2} = 3.5 \]Multiply through by \( v_2(v_2 - 10) \) to clear the denominators: \[ 60v_2 + 60(v_2 - 10) = 3.5v_2(v_2 - 10) \]

Step 4: Expand and Rearrange Equation

Expand and simplify the equation: \[ 60v_2 + 60v_2 - 600 = 3.5v_2^2 - 35v_2 \]Combine and rearrange terms to obtain a quadratic equation: \[ 3.5v_2^2 - 155v_2 + 600 = 0 \]

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( v_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3.5 \), \( b = -155 \), \( c = 600 \). Calculate:1. \( b^2 - 4ac = (-155)^2 - 4 \cdot 3.5 \cdot 600 \)2. Compute the discriminant and solve for \( v_2 \):Finally find acceptable root for \( v_2 \) and calculate corresponding \( v_1 = v_2 - 10 \).

Step 6: Verify the Solution

Compute the total travel time using the speeds found: \[ \frac{60}{v_1} + \frac{60}{v_2} \] should equal 3.5 hours as given, to validate the speeds.

Key Concepts

Quadratic equations

A quadratic equation can help solve problems where unknowns are squared. Commonly, it takes the form \( ax^2 + bx + c = 0 \). This means it features terms squared (\( x^2 \)), linear (\( x \)), and constant (\( c \)). In our word problem, the quadratic equation emerges from the speed and distance relationship.
This relationship becomes complex when accounting for varying speeds, like in round trips. Thus, we derive a formula involving squared variables. To resolve this, we use the quadratic formula:

• \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides solutions by evaluating possible values that \( v_2 \) can take. It's essential to simplify the square root term first (the discriminant \( b^2 - 4ac \)), ensuring real solutions exist if the discriminant is zero or positive. Avoid solutions yielding negative speeds or those outside practical bounds.

Variable substitution

In algebra, variable substitution replaces variables with expressions or other variables, simplifying equations. In this problem, we substitute \( v_1 = v_2 - 10 \) since the speed going is 10 km/h less than the speed returning.
Substitution helps us express the relationship between the two variables, \( v_1 \) and \( v_2 \), reducing the number of unknowns. By substituting, we transform one equation into another easier or critical part of solving our equation efficiently.

• Helps transform complicated equations into simpler forms
• Reduces the need to work with multiple unknowns at once

To apply variable substitution, identify given relationships and ensure the expressions used make the final equation simpler to solve. This strategic approach clarifies and simplifies problems, especially useful in speed and distance calculations.

Speed and distance calculations

Understanding speed and distance calculations can often involve forming equations based on travel criteria. The key formula used is \( \text{time} = \frac{\text{distance}}{\text{speed}} \). For a round-trip, where parameters can vary for each direction, we create separate time equations for each leg of the journey.
This exercise illustrates combining both travel times to equal the total travel time:

• Time going: \( \frac{60}{v_1} \)
• Time returning: \( \frac{60}{v_2} \)

Both contribute to the total time of 3.5 hours.To solve:

• Establish the individual time equations using the distance formula
• Add these times for a complete journey equation
• Use observations about speed differences to express various speeds

These fundamentals, when paired with algebraic principles, allow for solving complex distance and time relationships effectively.