Good evening Annela! Here is the relevant velocity time graph of the girl's motion.

Of course, you will need to know how the graph of y = e^-x looks like, since it is on your syllabus. However, for the time being, I will plot them for you to have a better idea of the question.

The red graph is the graph of v = -21/2 * e^(-t/80) - 2 which is basically the velocity graph of motion.

The girl stops cycling when t = 1.5 m/s, indicated by the blue horizontal line.

The girl stops cycling when t = 80 ln 3 (I have drawn a green vertical line to show the speed when t = 80 ln 3).

As we can see, the velocity v never gets to zero before t = 80 ln 3 (it does, however, reach zero and go negative at a much later time). So in this case the girl always travels in the "forward" direction, which is up the hill. There is no need to worry about the change in directions in this case, so we can just use t = 80 ln 3 and t = 0 as our upper and lower limits respectively.

Of course, the girl is initially 500 m from the top of the hill (i.e. the girl is at a height 500 m below this) and the girl cycles UP the hill towards the top. In doing so, the distance to the top of the hill gets shortened by 1 metre for every 1 metre moved upwards.

We find that the girl cycles 384 m upwards, so the girl is now 116 m away from the top of the hill. This is the distance the girl needs to walk to reach the top of the hill.

Let me know if you need more explanation and I will do my best to explain them in even greater detail.