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Shawn's engineering club is taking a trip by bus to a robotics fair. The fair is 175 miles from Shawn's school. The bus driver...Asked by Imtherizzler_
Shawn's engineering club is taking a trip by bus to a robotics fair. The fair is 175 miles from Shawn's school. The bus driver plans to stop and refuel after 2 hours, then complete the trip. The bus driver plans to drive at an average speed of 50 miles per hour. How many hours will the second part of the trip take?
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Answered by
GPT-4o mini
To determine how many hours the second part of the trip will take, we first calculate how far the bus will travel in the first 2 hours.
1. **Calculate distance traveled in the first 2 hours**:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 50 \, \text{miles/hour} \times 2 \, \text{hours} = 100 \, \text{miles}
\]
2. **Calculate the remaining distance to the fair**:
The total distance to the fair is 175 miles. After traveling 100 miles, the remaining distance is:
\[
\text{Remaining distance} = 175 \, \text{miles} - 100 \, \text{miles} = 75 \, \text{miles}
\]
3. **Calculate the time needed to cover the remaining distance**:
We use the average speed to find the time required to travel the remaining 75 miles:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{75 \, \text{miles}}{50 \, \text{miles/hour}} = 1.5 \, \text{hours}
\]
Thus, the second part of the trip will take \( \boxed{1.5} \) hours.
1. **Calculate distance traveled in the first 2 hours**:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 50 \, \text{miles/hour} \times 2 \, \text{hours} = 100 \, \text{miles}
\]
2. **Calculate the remaining distance to the fair**:
The total distance to the fair is 175 miles. After traveling 100 miles, the remaining distance is:
\[
\text{Remaining distance} = 175 \, \text{miles} - 100 \, \text{miles} = 75 \, \text{miles}
\]
3. **Calculate the time needed to cover the remaining distance**:
We use the average speed to find the time required to travel the remaining 75 miles:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{75 \, \text{miles}}{50 \, \text{miles/hour}} = 1.5 \, \text{hours}
\]
Thus, the second part of the trip will take \( \boxed{1.5} \) hours.
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