A boat traveled upstream 90 miles at an average speed of \((\mathrm{v}-3)\) miles per hour and then traveled the same...

1. Translate the problem requirements: A boat travels 90 miles upstream at speed \((\mathrm{v}-3)\) mph and 90 miles downstream at speed \((\mathrm{v}+3)\) mph. The upstream trip takes 0.5 hours longer than downstream. We need to find the downstream travel time in hours.
2. Set up time relationships using distance-speed-time: Express upstream and downstream times in terms of the variable v, recognizing that time = distance ÷ speed.
3. Create equation from the time difference constraint: Use the fact that upstream time minus downstream time equals 0.5 hours to form a solvable equation.
4. Solve for the boat's speed in still water: Algebraically manipulate the equation to find the value of v.
5. Calculate the downstream travel time: Substitute the found value of v into the downstream time formula to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in this problem using everyday language first.

Imagine you're in a boat on a river. When you go upstream (against the current), the water pushes back against you, making you slower. When you go downstream (with the current), the water helps push you forward, making you faster.

In our problem:

• The boat travels 90 miles upstream at a speed of \((\mathrm{v}-3)\) mph
• The boat travels 90 miles downstream at a speed of \((\mathrm{v}+3)\) mph
• Here, 'v' represents the boat's speed in still water, and '3' represents the current speed
• The upstream trip takes 0.5 hours (30 minutes) longer than the downstream trip
• We need to find how long the downstream trip took

Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships

2. Set up time relationships using distance-speed-time

Now let's think about the relationship between distance, speed, and time in plain English.

When you travel a certain distance at a certain speed, the time it takes equals the distance divided by the speed. It's like asking: "If I need to go 60 miles at 30 mph, how long will it take?" The answer is \(60 ÷ 30 = 2\) hours.

For our boat problem:

• Time upstream = Distance ÷ Speed upstream = \(\frac{90}{\mathrm{v}-3}\) hours
• Time downstream = Distance ÷ Speed downstream = \(\frac{90}{\mathrm{v}+3}\) hours

Notice how the upstream time will be larger because \((\mathrm{v}-3)\) is smaller than \((\mathrm{v}+3)\), which makes sense since going against the current takes longer.

3. Create equation from the time difference constraint

The problem tells us that the upstream trip takes half an hour longer than the downstream trip. In mathematical terms:

Time upstream - Time downstream = 0.5 hours

Substituting our expressions:

\(\frac{90}{\mathrm{v}-3} - \frac{90}{\mathrm{v}+3} = 0.5\)

This equation captures the key relationship in the problem. The left side represents the difference in travel times, and the right side is the given half-hour difference.

4. Solve for the boat's speed in still water

Now we need to solve this equation step by step. Let's work with the equation:

\(\frac{90}{\mathrm{v}-3} - \frac{90}{\mathrm{v}+3} = 0.5\)

To solve this, we'll find a common denominator for the fractions on the left side:

\(\frac{90(\mathrm{v}+3) - 90(\mathrm{v}-3)}{(\mathrm{v}-3)(\mathrm{v}+3)} = 0.5\)

Simplifying the numerator:

\(90(\mathrm{v}+3) - 90(\mathrm{v}-3) = 90\mathrm{v} + 270 - 90\mathrm{v} + 270 = 540\)

Simplifying the denominator:

\((\mathrm{v}-3)(\mathrm{v}+3) = \mathrm{v}² - 9\)

So our equation becomes:

\(\frac{540}{\mathrm{v}² - 9} = 0.5\)

Cross multiply:

\(540 = 0.5(\mathrm{v}² - 9)\)
\(540 = 0.5\mathrm{v}² - 4.5\)
\(544.5 = 0.5\mathrm{v}²\)
\(\mathrm{v}² = 1089\)
\(\mathrm{v} = 33\)

Process Skill: MANIPULATE - Algebraic manipulation to isolate the variable

5. Calculate the downstream travel time

Now that we know \(\mathrm{v} = 33\), we can find the downstream travel time.

Downstream speed = \(\mathrm{v} + 3 = 33 + 3 = 36\) mph
Downstream time = Distance ÷ Speed = \(90 ÷ 36 = 2.5\) hours

Let's verify this makes sense:

• Upstream speed = \(33 - 3 = 30\) mph
• Upstream time = \(90 ÷ 30 = 3\) hours
• Time difference = \(3 - 2.5 = 0.5\) hours ✓

4. Final Answer

The boat traveled downstream for 2.5 hours.

This matches answer choice A. 2.5.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the direction of speed effects

Students often confuse which direction gets which speed. They might think upstream speed is \((\mathrm{v}+3)\) and downstream speed is \((\mathrm{v}-3)\), reversing the logic. Remember: upstream means going AGAINST the current (so the current slows you down, hence \(\mathrm{v}-3\)), while downstream means going WITH the current (so the current speeds you up, hence \(\mathrm{v}+3\)).

2. Incorrectly setting up the time difference equation

Students frequently write the time difference equation backwards as: Time downstream - Time upstream = 0.5, instead of the correct Time upstream - Time downstream = 0.5. Since upstream takes longer, it should be the larger value being subtracted from.

3. Confusing what the question is asking for

The question asks for downstream travel time, but students might set up their entire approach to find upstream time, the speed v, or the time difference. Always double-check what the question is specifically asking for before you start solving.

Errors while executing the approach

1. Algebraic errors when finding common denominator

When simplifying \(90(\mathrm{v}+3) - 90(\mathrm{v}-3)\), students often make sign errors and get \(90\mathrm{v} + 270 - 90\mathrm{v} - 270 = 0\) instead of the correct \(90\mathrm{v} + 270 - 90\mathrm{v} + 270 = 540\). Pay careful attention to the negative sign in front of \(90(\mathrm{v}-3)\) and how it affects the signs when distributing.

2. Arithmetic mistakes in cross multiplication

When cross multiplying \(\frac{540}{\mathrm{v}² - 9} = 0.5\), students might incorrectly calculate \(540 = 0.5(\mathrm{v}² - 9)\) as \(540 = 0.5\mathrm{v}² - 9\) instead of \(540 = 0.5\mathrm{v}² - 4.5\). Remember that \(0.5 × (-9) = -4.5\), not -9.

3. Taking the wrong square root

When solving \(\mathrm{v}² = 1089\), students might calculate the square root incorrectly or take the negative value. Since v represents speed, it must be positive, so \(\mathrm{v} = 33\), not \(\mathrm{v} = -33\).

Errors while selecting the answer

1. Providing the upstream time instead of downstream time

After calculating both upstream time (3 hours) and downstream time (2.5 hours), students might select 3 hours because they spent more time calculating it or because it seems like the 'main' answer. Always refer back to what the question specifically asks for.

2. Giving the boat's still-water speed instead of the time

Since we calculated \(\mathrm{v} = 33\) mph during the solution process, students might mistakenly think this is the final answer. Remember that the question asks for time in hours, not speed in mph.