Problem 48 - Precalculus: Mathematical for Calculus

Question

A boat on a river travels downstream between two points, \(20 \mathrm{mi}\) apart, in one hour. The return trip against the current takes \(2 \frac{1}{2}\) hours. What is the boat's speed, and how fast does the current in the river flow?

Step-by-Step Solution

Answer

The boat's speed in still water is 14 mph, and the current's speed is 6 mph.

Step 1: Understand the Problem

We need to find two things: the speed of the boat in still water and the speed of the river current. We have information on the distance between two points (20 miles) and the time it takes for the boat to travel downstream (1 hour) and upstream (2.5 hours).

Step 2: Define Variables

Let the speed of the boat in still water be \(b\) miles per hour and the speed of the current be \(c\) miles per hour. When going downstream, the boat travels with the current, so its effective speed is \(b + c\). When going upstream, the boat travels against the current, so its effective speed is \(b - c\).

Step 3: Set Up Equations for Downstream and Upstream

Using the relation \(\text{distance} = \text{speed} \times \text{time}\): For downstream: \(b + c = \frac{20}{1} = 20\).For upstream: \(b - c = \frac{20}{2.5} = 8\).

Step 4: Solve the System of Equations

We have the system of equations:1. \(b + c = 20\)2. \(b - c = 8\)Add these two equations to eliminate \(c\):\(2b = 28\)\(b = 14\). Now substitute back to find \(c\): \(14 + c = 20\)\(c = 6\).

Step 5: Verify the Solution

The boat's speed in still water is 14 mph, and the current's speed is 6 mph. For downstream: the effective speed is 14 + 6 = 20 mph, taking 1 hour to travel 20 miles. For upstream: the effective speed is 14 - 6 = 8 mph, taking 2.5 hours to travel 20 miles, both consistent with the problem statement.

Key Concepts

Systems of Equations

Systems of equations are sets of equations with multiple variables that are solved together to find a common solution. In our boat and river problem, we need to determine two unknowns: the speed of the boat in still water, denoted as \(b\), and the speed of the river current, denoted as \(c\).
By setting up a system of equations, we can simultaneously solve for both these variables. Each equation represents a different condition described in the original problem.

• When the boat travels downstream, the equation is \(b + c = 20\). This means the effective speed is the sum of the boat's speed and the current's speed.
• When the boat goes upstream, the equation becomes \(b - c = 8\). This means the effective speed is the boat's speed minus the current's speed.

To solve these equations together effectively, we use methods such as addition or substitution. By adding the equations, we can eliminate one of the variables, simplifying the process to find the unknowns.

Distance-Speed-Time Relationships

The relationship between distance, speed, and time is fundamental in solving many physics and math problems. The formula \(\text{distance} = \text{speed} \times \text{time}\) is essential to understand and apply. In the context of our boat problem, this formula allows us to set up the initial equations which we use for solving later.
For example:

• Going downstream: The boat covers 20 miles in 1 hour, yielding the equation \(b + c = 20\)
• Returning upstream: The same distance is covered in 2.5 hours, leading to the equation \(b - c = 8\)

By considering these relationships, we can accurately derive the speed equations that define the motion of the boat both with and against the current. The given times and distances provide a concrete basis to solve for unknown speeds.

Algebraic Manipulation

Algebraic manipulation involves the process of rearranging and solving equations to isolate unknown variables. To find the boat's speed and the river current's speed, we utilized algebraic techniques on our two equations.
By following these steps:

• We start with two equations: \(b + c = 20\) and \(b - c = 8\).
• Add these equations to eliminate \(c\), giving \(2b = 28\).
• Solve for \(b\): dividing both sides by 2, we find \(b = 14\).
• Substitute \(b\) back into one of the original equations to find \(c\): \(14 + c = 20\), so \(c = 6\).

These steps demonstrate simple but powerful strategies to handle equations with multiple variables. Successful manipulation allows us to solve a system of equations easily, obtaining solutions that satisfy all given conditions.