The Train and the Bicycle
A student lives 12 km from school. One morning she walks to the station at 4 km/h, waits 6 minutes for a train, then rides the train at 40 km/h to the station nearest school, and walks the final 0.5 km at 4 km/h. The whole trip takes 42 minutes. The student wonders whether she could instead cycle the entire 12 km at 15 km/h and arrive sooner. She also wants to know at what distance from home the train portion begins, assuming the first walking segment and the final walking segment are at the same speed.
1. Using the given total time of 42 minutes, find the distance she walks to the station. Set up an equation in hours.
Read the worked answer
Let the first walking distance be $d$ km. Then the train distance is $12 - d - 0.5 = 11.5 - d$ km.
Times in hours:
- First walk: $\dfrac{d}{4}$
- Wait: $\dfrac{6}{60} = 0.1$
- Train: $\dfrac{11.5 - d}{40}$
- Final walk: $\dfrac{0.5}{4} = 0.125$
Total $= \dfrac{42}{60} = 0.7$ hours.
$$\frac{d}{4} + 0.1 + \frac{11.5 - d}{40} + 0.125 = 0.7$$
Multiply by 40: $10d + 4 + 11.5 - d + 5 = 28$.
$$9d + 20.5 = 28 \implies 9d = 7.5 \implies d = \frac{5}{6} \approx 0.833 \text{ km}$$
She walks about 0.83 km to the station.
2. Compare the total time of the train trip (42 minutes) with cycling the whole 12 km at 15 km/h. Which is faster, and by how much?
Read the worked answer
Cycling time: $\dfrac{12}{15} = 0.8$ hours $= 48$ minutes.
Train trip: 42 minutes.
Difference: $48 - 42 = 6$ minutes.
The train trip is faster by 6 minutes. The student's guess that cycling arrives sooner is incorrect.
3. The student says, 'If I could remove the 6-minute wait, the train trip would be under 40 minutes.' Check this claim and state the new total time.
Read the worked answer
Current total: 42 minutes, which includes the 6-minute wait.
Removing the wait: $42 - 6 = 36$ minutes.
$36 < 40$, so the claim is correct.
Without the wait, the trip takes 36 minutes.