Departure: | Cleveland, OH |
Arrival: | Martinsburg, WV |
Fastest route: | 9h 7min |
Distance: | 447km |
Cheapest route: | $50 |
Transfers: | 0 |
Train companies: | Amtrak |
One Passenger / One Trip
1:54am
Cleveland, OH
Cleveland
11:01am
Martinsburg, WV
Martinsburg
9h 7min
Amtrak
Capitol Limited
$50
The Cleveland - Martinsburg route has approximately 1 frequencies and its minimum duration is around 9h 7min. It is important you book your ticket in advance to avoid running out, since $50 tickets tend to run out quickly.
The distance between Cleveland and Martinsburg is around 447 kilometers and bus companies that can help you in your journey are: Amtrak.
The train journey may vary depending on the stops. The minimum duration is usually around 9h 7min to cover 447 kilometers.
According to our data, the cheapest ticket costs $50 and leaves Cleveland. You will not have to do any transfers, the trip will go direct to Martinsburg.
The last train leaves at 1:54am from Cleveland and arrives at 11:01am at Martinsburg. It will take 9h 7min, its price is $50 and the number of changes will be 0.
Yes, there are direct train routes, their duration is usually around 9h 7min and the price is $50.