Although Patrickt is quoting the laws correctly he may be incorrectly picking which variables are constant. An electric motor is turning a load and the torque required to turn that load is not changing. If the wire is longer and thus has more resistance, there will be more voltage drop across the wire (IR drop). So less voltage gets to the motor. If the battery can supply more current, which it can, the motor will pull more current to make the power needed to turn the load.
On AC motors what I have said is absolutely 100% correct because motor speed is a factor of the Hz. When voltage dips in a plant the ac motors pull more current because the work or wattage is constant. W = E * I so if W is constant and E goes down then I must go up. Now with that said DC is a little different because Motor speed is a function of the voltage. So when voltage goes down the motor rpm will be less and the motor will be doing less work. Now with both the work and the voltage changing it is more difficult to predict what the current will do, as it depend on how much less power is needed to turn the motor at the lower speed. Since the voltage drop in the wire is going to be slight, I suspect the work to turn the motor will drop an insignificant amount relative to the voltage drop and the motor will pull more amps.
PS
On second thought. On a dc motor current is proportional to the torque the electric motor is putting out. In the rpm range change of the voltage drop in the wire, the torque required to turn the gasoline engine is going to be negligible, although theoretically it would be less. Therefore, the current draw will remain unchanged. Other variables such as
oil temp and compression will be bigger than the voltage drop in the wire. However if the voltage drop in the wire were to be large enough to substantially reduce the rpm of the motor and cranking engine, the torque require to rotate the engine would go down and the dc motor would pull less amps, because the torque load was reduced.