Rate of Travel
Date: 05/15/2003 at 09:04:48 From: Wendy Subject: Rate of travel A passenger on a train traveling at 135 km/h walks toward the back of the train at a rate of 7 km/h. What is the passenger's rate of travel with respect to the ground? What does "with respect to the ground" mean? I think it is a trick question because the person is walking 7 km/h no matter what the speed of the train.
Date: 05/15/2003 at 14:38:20 From: Doctor Ian Subject: Re: Rate of travel Hi Wendy, Suppose there is a flatcar being pulled along at 5 mph, with two guys riding on it: \A/ \B/ | | / \ / \ --------------------- -- 5 mph --> o o There is another guy standing by the tracks watching: \A/ \B/ | | / \ / \ --------------------- -- 5 mph --> o o C /|\ / \ Now, 5 miles per hour is the same as 88 feet per second, which is the same as 22 feet every 15 seconds. So if we wait 15 seconds, the situation looks like this: \A/ \B/ | | / \ / \ --------------------- -- 5 mph --> o o C /|\ / \ That is, both guys on the train have moved 22 feet as seen by the guy on the ground. However, suppose that as the train moves, guy B starts walking toward guy A at a speed of 5 mph. Now, if we wait 15 seconds, the situation looks like this: \A/ \B/ | | / \ / \ --------------------- -- 5 mph --> o o C /|\ / \ That is, from the point of view of guy A, guy B has been moving at 5 miles per hour. But from the point of view of guy C, guy B hasn't moved at all. He's still in the same place he was before, even though the _train_ has moved. Does this make sense? The language we use to describe this situation is that guy B has moved at 5 miles per hour 'with respect to' guy A (i.e., as seen from guy A's point of view), and that guy B has moved at 0 miles per hour 'with respect to' guy C. Since guy C is stationary on the ground, we also say that guy B has moved 0 mph with respect to the ground. And since guy A is stationary on the train, we say that guy A has moved 0 mph with respect to the train, but 5 mph with respect to the ground. In many (perhaps most) cases, we leave the observer unspecified. If a policeman gives you a ticket for driving 80 miles per hour, he means 'with respect to the ground', even though he won't specify that. (If he's right behind you at 80 mph when he turns his siren on, you're not moving at all with respect to him!) Now, suppose we go back to our train, but this time, instead of walking at 5 mph, guy B walks toward guy A at 3 mph with respect to guy A. What does guy C see? Well, imagine that there is a fence blocking his view of the train, so all he can see is the heads of guys A and B. What guy C will see is guy A moving to the right at 5 mph, and guy B moving to the right at 2 mph (i.e., 5 mph minus 3 mph). The point of all this is that whenever you say that something is moving at a certain speed, it implies that the speed is a speed that would be measured BY SOME OBSERVER. This is necessary, because as we see, different observers will end up making different measurements. Note that by thinking in terms of relative speeds, certain kinds of problems can be made much easier. For example, suppose two cars start a race at the same time, around a track that is 1 1/2 miles long, and one moves at 151 mph while the other moves at 154 mph. How long will it take the faster car to lap the slower one? If you try to solve this using their speeds relative to the ground, it's a mess. But you can reason this way: the faster car is moving at 3 mph WITH RESPECT TO the slower car. This is true whether their speeds with respect to the ground are 151 and 154 mph, or 18 and 15 mph, or 46.3 and 49.3 mph. And to lap the slower car, the faster car needs to get ahead by the length of the track. In other words, we want to know how long it takes to go 1 1/2 miles at 3 mph. And that's an easy problem to solve: it takes half an hour, or 30 minutes. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum