Problem 24 - Precalculus: Functions and Graphs

Question

A man rows a boat 500 feet upstream against a constant current in 10 minutes. He then rows 300 feet downstream (with the same current) in 5 minutes. Find the speed of the current and the equivalent rate at which he can row in still water.

Step-by-Step Solution

Answer

The speed of the current is 5 feet per minute, and the man's rowing speed in still water is 55 feet per minute.

Step 1: Define Variables

Let's define some variables to make the problem clearer. Let \( r \) be the rate at which the man rows in still water (in feet per minute), and let \( c \) be the speed of the current (in feet per minute). We need to find these two values.

Step 2: Equation for Upstream Rowing

When rowing upstream, the effective speed of the boat is \( r - c \) because the current is against the motion. The man rows 500 feet upstream in 10 minutes, so the equation is: \[ 10(r - c) = 500. \] Simplifying this, we get \( r - c = 50. \)

Step 3: Equation for Downstream Rowing

When rowing downstream, the effective speed of the boat is \( r + c \) because the current aids in the motion. The man rows 300 feet downstream in 5 minutes, so the equation is: \[ 5(r + c) = 300. \] Simplifying this, we get \( r + c = 60. \)

Step 4: Solve the System of Equations

Now, we have two equations: \( r - c = 50 \) and \( r + c = 60. \) We can add these equations to eliminate \( c \): \[ (r - c) + (r + c) = 50 + 60, \] which simplifies to \( 2r = 110 \), therefore \( r = 55. \) Then, substitute \( r = 55 \) back into one of the equations, say \( r - c = 50 \): \[ 55 - c = 50 \] which gives \( c = 5. \)

Key Concepts

Current Speed Calculation

To find the current's speed in this problem, we must analyze how it affects the rower's journey in both directions.
The current's speed, denoted as \( c \), creates different effective speeds when rowing upstream and downstream.

• Upstream: The current works against the rower.
• Downstream: The current aids the rower.

When rowing upstream, the effective speed is the rower's speed minus the current's speed: \( r - c \). The problem states that this rowing takes 10 minutes to cover 500 feet, leading to the equation: \[10(r - c) = 500.\]After simplification, the impact of the current gives \( r - c = 50 \).
Similarly, downstream, the effective speed increases due to the current: \( r + c \). Here, 300 feet are covered in 5 minutes, with the equation:\[5(r + c) = 300.\]This simplifies to \( r + c = 60 \).
Computing these, we find the current's speed is 5 feet per minute.

Rowing in Still Water

Understanding rowing in still water means focusing purely on the rower's ability, unaffected by current forces.
This rate, represented by \( r \), reflects the rower's natural speed without any assistance or hindrance.

• When currents are present, they modify the effective speed.
• Two main contexts: moving against and with the current.

We use these real-world rowing scenarios to isolate how fast the rower would travel in ideal conditions.
By solving: \[ r - c = 50 \]\[ r + c = 60 \]By eliminating the variable \( c \), we sum both simplified equations, resulting in:\[ 2r = 110 \]and finding:\[ r = 55 \] feet per minute.
So, the rower would travel at 55 feet per minute in still water.

System of Equations

A system of equations, like the one in this problem, uses multiple equations to find unknown values.
Here, the purpose is to find both the rate of rowing in still water \( r \) and the current speed \( c \).

• Two equations were derived from rowing times and conditions.
• The technique of addition helps simplify and address two equations simultaneously.

The equations formed were based on the conditions:\[ r - c = 50 \]and\[ r + c = 60 \]By adding them, we can solve directly for \( r \):\[ 2r = 110 \]Solving for \( r \) gives the rowing speed in still water:\[ r = 55 \textrm{ feet per minute}\]Then, substituting \( r = 55 \) back into either equation, the current speed \( c \) can be calculated, specifically:\[ 55 - c = 50\]giving:\[ c = 5 \textrm{ feet per minute}\]Using simple arithmetic, this approach elegantly deciphers each variable's value, illustrating the power of solving systems of equations.