Departure: | Martinsburg, WV |
Arrival: | Cleveland, OH |
Fastest route: | 9h 8min |
Distance: | 447km |
Cheapest route: | $50 |
Transfers: | 0 |
Train companies: | Amtrak |
One Passenger / One Trip
5:45pm
Martinsburg, WV
Martinsburg
2:53am
Cleveland, OH
Cleveland
9h 8min
Amtrak
Capitol Limited
$50
The Martinsburg - Cleveland route has approximately 1 frequencies and its minimum duration is around 9h 8min. It is important you book your ticket in advance to avoid running out, since $50 tickets tend to run out quickly.
The distance between Martinsburg and Cleveland 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 8min to cover 447 kilometers.
According to our data, the cheapest ticket costs $50 and leaves Martinsburg. You will not have to do any transfers, the trip will go direct to Cleveland.
The last train leaves at 5:45pm from Martinsburg and arrives at 2:53am at Cleveland. It will take 9h 8min, its price is $50 and the number of changes will be 0.
Yes, there are direct train routes, their duration is usually around 9h 8min and the price is $50.