To find the distance Fritz travels to work, let's break this down using the formula for distance:

$\text{Distance} = \text{Speed} \times \text{Time}$

Given:

• The driving time is 35 minutes.
• The train time is 15 minutes.
• The train's speed is 80 miles per hour faster than Fritz's driving speed.
• The distance traveled by both the train and the car is the same.

Let:

• $d$ be the distance Fritz travels to work.
• $v$ be Fritz's driving speed in miles per hour.

Step 1: Convert time to hours

Since speed is in miles per hour, we convert the time from minutes to hours:

• Driving time: $35 \, \text{minutes} = \frac{35}{60} \, \text{hours} = \frac{7}{12} \, \text{hours}$
• Train time: $15 \, \text{minutes} = \frac{15}{60} \, \text{hours} = \frac{1}{4} \, \text{hours}$

Step 2: Express distance for both cases

1. For driving: $d = v \times \frac{7}{12}$

2. For the train: Since the train travels 80 miles per hour faster than Fritz drives, the train's speed is $v + 80$. The distance for the train is: $d = (v + 80) \times \frac{1}{4}$

Step 3: Set the distances equal

Since the distance is the same for both, we set the two equations equal to each other: $v \times \frac{7}{12} = (v + 80) \times \frac{1}{4}$

Step 4: Solve for $v$

First, multiply both sides of the equation to eliminate the fractions: $12 \times \left( v \times \frac{7}{12} \right) = 12 \times \left( (v + 80) \times \frac{1}{4} \right)$ $7v = 3(v + 80)$

Expand the right-hand side: $7v = 3v + 240$

Subtract $3v$ from both sides: $4v = 240$

Solve for $v$: $v = 60$

Step 5: Find the distance

Now that we know Fritz's driving speed is 60 miles per hour, we can use either the driving or train equation to find the distance. Using the driving equation: $d = 60 \times \frac{7}{12} = 35 \, \text{miles}$

Thus, the distance Fritz travels to work is 35 miles.

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Do you want any more details or have any questions? Here are 5 related questions to expand on this:

1. What would the driving time be if Fritz increased his speed by 10 mph?
2. How would the distance change if the train traveled at 100 mph faster instead of 80 mph?
3. If Fritz took a route that was 5 miles longer, how would that affect his driving time?
4. How could you calculate the average speed of both trips combined?
5. What if the driving time was reduced to 30 minutes, but the train speed stayed the same?

Tip: Always convert time to the same units as speed (e.g., hours for mph) when using the distance formula.