Alternatively, we could choose a frame of reference S′ situated in the first car. In this case, the first car is stationary and the second car is approaching from behind at a speed of v 2 − v 1 = 8 m/s. To catch up to the first car, it will take a time of ⁠ d / v 2 − v 1 ⁠ = ⁠ 200 / 8 ⁠ s, that is, 25 seconds, as before. Note how much easier the problem becomes by choosing a suitable frame of reference. The third possible frame of reference would be attached to the second car. That example resembles the case just discussed, except the second car is stationary and the first car moves backward towards it at 8 m/s.