Since the ships are moving with the same velocity in S′ before acceleration, the initial rest length L in S is shortened in S′ by L o l d ′ = L / γ γ due to length contraction. From the frame of S′, B starts accelerating before A and also stops accelerating before A. Due to this B will always have higher velocity than A up until the moment A is finished accelerating too, and both of them are at rest with respect to S′. The distance between B and A keeps on increasing till A stops accelerating. Although A's acceleration timeline is delayed by an offset of Δ Δ t ′ , both A and B cover the same distance in their respective accelerations. But B's timeline contains acceleration and also being at rest in S` for Δ Δ t ′ till A stops accelerating. Hence the extra distance covered by B during the entire course can be calculated by measuring the distance traveled by B during this phase. Dewan arrived at the relation (in different notation):