Going on a Trip!

The Trip physics question is both interesting and annoying! The question goes like this; "A motorist wished to travel 40 kilometers at an average speed of 40km/h. During the first 20 kilometers, an average speed of 40km/h is maintained. During the next 10 kilometers, however, the motorist goofs off and averages only 20 km/h. To drive the last 10 kilometers and average 40 km/h the motorist mist drive: a.)60km/h b.) 80km/h c.) 90km/h or d.) faster than the speed of light? My initial guess at the problem was 80 km/h since the motorist drove half of the average speed. Since the motorist drove at a speed of 20km/h for 10 km I assumed that she would have to drive 80 km/h to make up for lost time. This answer was incorrect. Although I was on the right path, I wasn't seeing the full picture. I missed the aspect of time and began to re-think my answer. I used the equation speed=distance/time. I rearranged this equation to discover the time and was shocked at my discovery. Since the motorist drove 40km/h for 20km the motorist drove for 30 minutes. I then discovered that the motorist drove at 20km/h for 10km indicating another 30 minute time interval. The motorist traveled 30km in an hour and initially wanted to travel 40km in one hour. This means that the motorist has to travel 10km instantaneously, faster than the speed of light. The speed formula can be manipulated to derive either the distance or the time and object has traveled when given the speed, distance or time. In this case you would rearrange the speed formula to time = distance/speed. I approached this problem trying to discover the average speed and was overlooking the time aspect. This problem has taught me to factor ALL variables and pay attention to little details in the equation. I will remember to underline all the variables discussed in the question in future problems.