Question

Finding the Current of a Stream Pamela requires 3 hours to swim 15 miles downstream on the Illinois River. The return trip upstream takes 5 hours. Find Pamela's average speed in still water. How fast is the current? (Assume that Pamela's speed is the same in each direction.

Short answer

Pamela's speed in still water is 4 mph. The current speed is 1 mph.

Step-by-step solution

- Define the Variables

Let’s denote Pamela’s speed in still water as \(v\) (in miles per hour), and the speed of the current as \(c\) (in miles per hour).

- Write the Equations for Downstream and Upstream

When Pamela swims downstream, her effective speed is \(v + c\) because the current aids her. The time taken downstream is 3 hours, so the equation is: \[15 = 3(v + c)\]

- Simplify the Downstream Equation

Solving for \(v + c\) in the downstream equation: \[ 15 = 3(v + c) \Rightarrow v + c = 5 \]

- Write the Equation for the Upstream Journey

When swimming upstream, Pamela’s effective speed is \(v - c\) because the current opposes her. The time taken upstream is 5 hours, so the equation is: \[15 = 5(v - c)\]

- Simplify the Upstream Equation

Solving for \(v - c\) in the upstream equation: \[ 15 = 5(v - c) \Rightarrow v - c = 3 \]

- Solve the System of Equations

Now, we have two equations: \ (1) \ v + c = 5 \ (2) \ v - c = 3\. Adding these equations together eliminates \(c\): \[ (v + c) + (v - c) = 5 + 3 \Rightarrow 2v = 8 \Rightarrow v = 4 \]

- Find the Speed of the Current

Substituting \(v = 4\) back into \(v + c = 5\): \[ 4 + c = 5 \Rightarrow c = 1 \]

Key concepts

Algebra

Algebra is a branch of mathematics that allows us to solve problems involving unknown values. It's like a puzzle where we find the missing pieces, often represented by letters like \(x\) and \(y\). In the context of our exercise, we used variables to represent Pamela's speed in still water (\(v\)) and the speed of the current (\(c\)). Algebra helps us form equations based on the information given in a problem. These equations contain variables and constants. We then use algebraic methods, like addition, subtraction, multiplication, and division, to manipulate the equations and solve for the unknowns.

Rate Problems

Rate problems typically involve finding the relationship between distance, time, and speed. They are a practical application of algebra. In this exercise, the rate problem is about understanding how the speed of the current affects Pamela's swimming speed. We know that:

• Downstream, the current helps Pamela swim faster.
• Upstream, the current slows her down.

To solve rate problems like this, we form equations based on the formula: \( \text{Rate} = \frac{\text{Distance}}{\text{Time}} \). Using this equation, we figured out the effective speeds:

• Downstream: \(v + c = 5 \text{ mph}\)
• Upstream: \(v - c = 3 \text{ mph}\)

Variables and Equations

Variables and equations are foundational concepts in algebra. A variable is a symbol (often a letter) used to represent an unknown value. An equation is a mathematical statement asserting the equality of two expressions. In our exercise, we defined variables:

• \(v\) for Pamela's speed in still water
• \(c\) for the current's speed

We created two equations based on the conditions:

• \(15 = 3(v + c)\) for the downstream trip
• \(15 = 5(v - c)\) for the upstream trip

We then simplified these equations and solved them simultaneously. By adding and subtracting the equations, we found the values of \(v\) and \(c\). This process is powerful because it allows us to find unknown values through logical steps.