8. A freight train leaves the town of Jenkinsville at 1:00 PM traveling due east at constant speed. Jim, a hobo, sneaks onto the train and falls asleep. At the same time, Julie leaves Jenkinsville on her bicycle, traveling along a straight road in a northeasterly direction (but not due northeast) at 10 miles per hour. At 1:12 PM, Jim rolls over in his sleep and falls from the train onto the side of the tracks. He wakes up and immediately begins walking at 3.5 miles per hour directly towards the road on which Julie is riding. Jim reaches the road at 2:12 PM, just as Julie is riding by. What is the speed of the train in miles per hour?
Solution: $\square$
Julie's distance is $(10 \mathrm{mph}) \cdot(6 / 5 \mathrm{hrs})=12$ miles. Jim's walking distance, after falling off the train, is $(3.5 \mathrm{mph}) \cdot(1 \mathrm{hr})=3.5$ miles at a right angle to the road. Therefore, Jim rode the train for $\sqrt{12^{2}+3.5^{2}}=\frac{1}{2} \sqrt{24^{2}+7^{2}}=25 / 2$ miles, and its speed is $(25 / 2 \mathrm{mi}) /(1 / 5 \mathrm{hr})=62.5 \mathrm{mph}$.