# Thread: boats and trains related rates

1. ## boats and trains related rates

A railroad bridge is 200 ft. above, and at right angles to a river. A man in a train, traveling at 20 mph, passes over the center of the bridge at the same instant a man in a motorboat, traveling 30 mph passes under the center of the bridge. How fast are the two men moving away from each other 10 seconds later?

Really could use some help solving this one...

2. Take the x-axis to be in the direction the man in the train is going, the y-axis in the direction the man in the boat is going, the z-axis up. Take the origin of the coordinate system to be the point where the man in the boat is as he passes under the bridge and take time t= 0, to be that moment. We are given their speed in miles per hour, but, since the bridge's height is given in feet and we are asked what happens 10 seconds later, it might be better to give them in feet per second. There are 5280 feet to a mile and (60)(60)= 3600 seconds per hour so 20 mph is (20)(5280)/3600)= 29 and 1/3 or 29.33 feet per second. 30 mph is 30(5280)/3600= 444 feet per second.

At t= 0, the man in the boat is at (0, 0, 0) and the man in the train is at (0, 0, 200).

Since the man in the boat is moving in the direction of the y-axis with speed 44 feet per second, in t seconds he will be at (0, 44t, 0). Since the man in the train is moving in the direction of the x-axis, with speed 29.33 feet per second, in t seconds he will be at (29.33t, 0, 200).

Use the distance formula to determine the distance between them after t seconds, differentiate that to determine how fast it is changing, and evaluate at t= 10.