MythBusters - velocity is relative

So, I complained about MythBuster's explanation of relative velocity. How would I explain this? I would start by saying that velocity is relative. Here is the definition for velocity:

i-8f40464ebeb959a679a87329c9a19b98-2010-04-09_la_te_xi_t_1.jpg

I put the "avg" in there because it is more true. If the acceleration is zero, I could drop this. For the rest of this post, I am going to assume zero acceleration. Ok. But what is the r vector? It is simply a vector from the origin to the object. Here is a picture.

i-7f20c0dbb4f8325143c0700293cee13e-2010-04-09_untitled_3.jpg

Simple, right? And so the velocity tells how this vector r changes. But wait. Who says that I used the correct origin? How do you know that is the correct one? Couldn't I use a different one? The origin is not a real thing, so I can change it if I like. What if there are two origins?

i-9a7d09fca8f9e661a6f50afb9f2a92ec-2010-04-09_untitled_4.jpg

The great thing is that it doesn't matter which coordinate system you use, both of the changes in position vectors (Delta r) are the same. What does this have to do with relative velocity? I will get there. What if one of my coordinate systems is moving with respect to the other. Notice I have to say "with respect to" because just like the ball's velocity is relative to the coordinate. To make this simple, I am going to only deal with 1-dimension (if you want 2-D relative velocity - here is a post on that).

Here is the deal. There will be one coordinate system (x) and another one (x') that is moving away from x with a speed s. The ball in question is only moving in the x (and x') direction. At t = 0, these two coordinates are at the same location (just for simplicity). Here is a picture at some time later (I call it t)

i-7b4f5945ae1a968fb739b68cd221a61d-2010-04-09_untitled_7.jpg

What is the connection between x2 and x'2? If the frame on the right is moving with a speed s with respect to the other frame then:

i-a0e477903fc21f2a9649c999243e272f-2010-04-09_la_te_xi_t_1_3.jpg

This form is only true if the two frames are at the same location at t = 0 - but in the end, the result will be the same no matter where the two frames are at t = 0.

Ok - now for the x-velocity in the first frame after this time t:

i-499d07940aac04db007b26ae6212fd07-2010-04-09_la_te_xi_t_1_4.jpg

I know an expression for x2 also, since the two frames where at the same place at t = 0, x1 and x'1 have the same value. This gives:

i-6cc6612aa8b357fa4a0e39fe3622d8a2-2010-04-09_la_te_xi_t_1_5.jpg

So, the velocity in the "stationary" frame is the velocity in the "moving" frame plus the speed of the moving frame with respect to the other frame.

Back to the Mythbusters

In the MythBusters episode, the stationary frame is the ground. The moving frame is the truck. They want to show that if you shoot a ball at -60 mph with respect to the truck (that would be v') and the truck is moving at 60 mph (that would be s) then v = 0 mph.

Other stuff

If you would like a homework assignment, you can show:

  • It doesn't matter that the two frames are not at the same place at t = 0
  • This works in 3-D as a vector equation.

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