| Departure: | Martinsburg, WV | 
| Arrival: | Cleveland, OH | 
| Fastest route: | 9h 8min | 
| Distance: | 447km | 
| Cheapest route: | $63 | 
| Transfers: | 0 | 
| Train companies: | Amtrak | 
One Passenger / One Trip
5:45pm
Martinsburg, WV
Martinsburg
2:53am
Cleveland, OH
Cleveland
9h 8min
Amtrak
Capitol Limited
$63
          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 $63 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 $63 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 $63 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 $63.