Twist on a familiar problem

We're all familiar with this problem:
Smith takes the train daily from his work in the city back to the suburb where he lives and is met at the station by his wife, who drives him home. One day Smith finishes his work earlier than usual and arrives in his suburb on hour earlier than usual. He then walks home from the station, and meets his wife on the way. The wife stops and takes Smith home; this brings Smith home 10 minutes earlier than usual. How long did Smith walk? The time for stopping and picking up is to be disregarded, and it is also assumed that the wife arrives at the station as the same moment as the train.

Here's the twist:
The road rises from the station to the house, so that the car travels more slowly from the station to the house than from the house to the station. We assume that the road rises equally sharply everywhere, so that the car travels equally fast everywhere from the house to the station, and similarly on its way back, but then at a lower speed.

Once again, Smith arrives in his suburb one hour earlier than usual, walks home from the station, and meets his wife. The latter, however, does not notice Smith, and drives on. "What a pity," Smith says to himself. "If my wife had seen me, I would have been home 18 minutes earlier than usual. I'll keep on walking; my wife will catch up with me in 21 minutes, and I'll be home at the usual time." How long had Smith been walking when his wife drove past him on her way to the station? The time the wife needs to ascertain that Smith is not on the train is to be disregarded; so it is assumed that after her arrival at the station his wife immediately returns to join Smith.