MI → MI · about 22 miles
Reed City to Stanwood bus schedule
Indian Trails runs 7 direct buses a week from Reed City to Stanwood. The fastest takes 35 minutes; every bus leaves at 12:45 PM. The standard fare is $7 one way.
Timetable data from · No schedule changes since at least
- Buses a week
- 7
- Fastest trip
- 35m
- Leaves at
- 12:45 PM
- Standard fare
- $7
A standard fare is the carrier's published one-way price. Many carriers price by demand, so the fare when you book can be higher or lower. Fares don't include any fees the carrier's ticket site may add; BusSchedules.com adds no fees.
Timetable
Every bus leaves from Reed City and arrives at Stanwood (Bauman's Party Store). Times are local to each stop. Timetables published by tds.ai, refreshed October 15, 2026. Schedules change, so check times when you book.
| Departs | Arrives | Duration | Days | Bus company | Ticket |
|---|---|---|---|---|---|
| 12:45 PM | 1:20 PM | 35m | Daily | Indian Trails Full timetable | Book (carrier website) |
Route map
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Where to board in Reed City Where you get off in Stanwood The route of the busiest trip
Where to catch the bus in Reed City
- Reed City Indian Trails bus stop Map of Reed City Indian Trails bus stop (Google Maps)
Where the bus stops in Stanwood
- Stanwood (Bauman's Party Store) Map of Stanwood (Bauman's Party Store) (Google Maps)
Return trip: Stanwood to Reed City
Questions
How long is the bus from Reed City to Stanwood?
The fastest direct bus takes 35 minutes.
How much is the bus from Reed City to Stanwood?
The standard one-way fare with Indian Trails is $7. Many carriers price by demand, so the fare when you book can be higher or lower. Fares don't include any fees the carrier's ticket site may add; BusSchedules.com adds no fees.
What time is the first bus from Reed City to Stanwood?
Every departure is at 12:45 PM (Daily).
Which bus companies go from Reed City to Stanwood?
Indian Trails runs direct buses on this route.
Where do I catch the bus in Reed City?
Buses to Stanwood leave from Reed City Indian Trails bus stop.