relative velocity

MythBusters - velocity is relative

rallain — Fri, 09 Apr 2010 20:45:26 +0000

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 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.

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?

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)

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:

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 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:

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.

rallain Fri, 04/09/2010 - 16:45

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frame of reference

mythbusters

Physics

MythBusters and physics terms

rallain — Thu, 08 Apr 2010 09:01:33 +0000

MythBusters and physics terms

In the last episode of MythBusters, they tried to reproduce the following experiment. Suppose you are driving in a car at 60 mph and you shoot a ball backwards at 60 mph (with respect to the car). Will the ball just drop (with respect to the ground)? Actually, it is a cool demo - I saw some Japanese show did this a while ago.

So, what is the problem? The problem is with the MythBusters' use of their terms to explain this thing. Let me look at a couple of the things they said to explain this (surprisingly, they described it several different ways). This first one is my favorite.

Bad physics 1

"it makes sense that an object subjected to equal and opposite forces would drop like a rock."

Where to begin? First, I am not sure where they get these equal and opposite forces. I would love for the science diagram to include a free body diagram for the ball after it is shot out of the truck. It would probably look something like this:

Oh, I know. Neither of those are real forces - but that must be what they are thinking. Now for the next problem with that statement. Here is a correct force diagram for an air plane flying with a constant velocity.

I am not going to talk about planes - this was just the first example I came up with. In this case the force pushing forward DOES equal the force pushing back. Does the plane drop like a rock? The idea that a net force of zero (vector) means no motion is what Aristotle would say.

Bad Physics 2

Here is another quote regarding the ball:

"Does the forward momentum and the backward momentum cancel out?"

I think the main error here is that they are using the term "momentum" to either mean velocity or force - not sure which. Also, if they are talking about momentum, the momentum of what? The truck and the ball?

Bad Physics 3

"...see if the energies cancel out"

Can you get energies to cancel? Well, you could get the potential plus kinetic to be zero joules, but would that count as canceling? You can never get two kinetic energies to add up to zero because 1) it isn't a vector and 2) it is always positive.

I usually don't attack the MythBusters, but I couldn't help it in this case. Dear MythBusters, if you want to explain some stuff in the future give me a call. I will gladly look over your stuff for you.

rallain Thu, 04/08/2010 - 05:01

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I saw only the final 1/3 of the episode, so didn't see what they claimed was the source of the "myth" (I did think their video was pretty cool).

I first heard this "myth" as it concerned the old B-58 Hustler bomber - capable of speeds up to mach 2.
The original design had a 20-mm gun in the tail, and the stories were that when the gun was fired during high-speed flight "the bullets simply left the barrels and fell to the ground". It was a cool visual for a kid - I have no idea whether it actually happened.

Since the bullets are shot at something that's presumably following the bomber at Mach 2, the impact is the same for the target even if the bullet doesn't move relative to the ground. But to open the big can of worms, what if you're leading the target by shooting slightly above it to make up for the bullet drop? If you're watching from the stationary observer point, do you see the bullet go up in a straight vertical line, followed by a reversal of directions (assuming the horizontal velocity vectors still cancel out)?

I think my problem was the approach as a whole - this is one of those problems that is better handled (mathematically) using frame of reference transformations rather than traditional Newtonian mechanics methods. In fact, in modern physics (for the undergraduate) this type of scenario is one of the first examples we're taught before going on to Einstein's Relativity and how light handles reference frames differently.

The first statement is clearly erroneous. It massacres both the first and third laws. The second statement is fine as all you have to do is multiply the velocity addition formula by mass to obtain a "momentum addition" formula:

m*v_ball_ground = m*v_ball_car + m*v_car_ground

The forward momentum of the car and the backwards momentum of the ball "cancel". The third statement could refer to work-energy for the ball:

K_final = K_initial + W

The work on the ball is negative and "cancels" the initial kinetic energy. I agree with the woman in the picture. Go pick on ESPN's "Sports Science" again because they clearly just make stuff up.

Yikes - Sounds like this could be a good exercise for high school physics students. "Find all of the mistakes Jamie & Adam make". This might also help the "everything that comes from the TV or teacher is right" syndrome.

@Erik,

I don't know if that is a guy or girl. But you are right, ESPN just makes up crap and I still love the MythBusters.

@Erik + Rhett,
That is a boy in drag. A loony one at that. The MythBusters are great, we've just got to remember that they are special-effects guys and not practicing physicists! Perhaps they could use your help after all Rhett...

Adam Savage and Jamie Hyneman are actually speaking at my campus tonight. Unfortunately I didn't get tickets or else I could explain to them what Newton's 3rd law means.

Hi Rhett,

I saw those issues in this episode and assumed you'd quickly blog about them, so I didn't. But I'm beginning to tire of Mythbusters, they're running woefully short of "myths" and are reaching.

The ball out of the truck was a great idea, it was a shame they did the force/energy/momentum talk. Then they say things like "science in progress." Well heck, science is always in progress. But they mean they're "doing science," why can't they take the time to learn the actual terminology?

@Rob,

The mythbusters are awesome at building stuff. Science, not so much. It used to be ok because they didn't try to explain science. In this case, I would have been ok if it had just been Tory and Grant talking - but when they put in the narrator, then it becomes official science explanation and at a different level.

The work on the ball is negative and "cancels" the initial kinetic energy. I agree with the woman in the picture. Go pick on ESPN's "Sports Science" again because they clearly just make stuff up. thank you good informayin good post

Actually this thread gives me an idea for a lesson. Teach students about airplanes and how they fly. (Any good ground school textbook will cover it nicely, let alone the classic stick and rudder). You can show how if you have more lift that gravity forces you climb, and the converse if you have less lift (which is why you throttle back to descend, and throttle up to ascend) You could also cover centripital forces with turning planes and how if you try to maintain level flight while turning depending on the angle you hold the wings at you will feel a g force. (Actually maybe get a hold of a flight simulator and demo it (except for the g forces, but when I took a few lessons it feeling the g forces pushed the concept home to me).

@Lyle,

One small (but important point) - if the lift is greater than gravity, you would _accelerate_ upwards. This could be still moving down though - think about doing a loop the loop - on the way down the plane is moving down but the acceleration is partly upward.

I do like the idea of the simulator though.

If you're watching from the stationary observer point, do you see the bullet go up in a straight vertical line, followed by a reversal of directions (assuming the horizontal velocity vectors still cancel out)?

But to open the big can of worms, what if you're leading the target by shooting slightly above it to make up for the bullet drop? If you're watching from the stationary observer point, do you see the bullet go up in a straight vertical line, followed by a reversal of directions (assuming the horizontal velocity vectors still cancel out)?

The gods of wind curse me (and average velocity)

rallain — Fri, 16 Jan 2009 16:52:55 +0000

The gods of wind curse me (and average velocity)

It has been windy here lately. Sometimes I think that is an ok thing. You see, when I ride my bike to work I am probably going to have the wind at my back for one of directions. It is great feeling like Lance Armstrong because of the boost you get from the wind. With a good wind at my back, I can almost keep up with the traffic (I would keep up if they went the 25 mph speed limit).

Of course, with a great boost comes a great drag. When I ride into the wind, I feel weak. I pedal as fast as I can and cars just whiz right by like I am standing still. When you are in a car, you don't really notice the wind. You notice it on a bike as though it were blowing in your face (because it is).

Yesterday it was windy. It seemed the god of the wind (Hermes? I don't know who you are, god of the wind - maybe that is why you smote me) was always in my face. It was no fun. When life gives you lemons, blog about it.

So, here is one of my favorite easy physics problems. Suppose I bike 2 miles to work. On the way there, the wind gods like me and I can go 20 mph. On the way back, I only go 10 mph. What is my average speed for the round trip?

First, notice that I said "average speed" and not "average velocity". The convention is that average velocity is defined as:

Where the vector r is the position of the object. If I take a round trip, my starting and ending position vectors are the same thing. This would make the average velocity zero (zero vector). Average speed, on the other hand, can be calculated as:

Where s is the total distance along the path (not zero for a round trip).

Ok, I want to calculate average speed. For this case, let me pretend that I am going all in one direction. So the problem would be that I travel 20 mph for 2 miles and then 10 mph for 2 miles. Now I can call my position along the x-axis.

Let me go ahead and give you the wrong answer. The wrong answer is that the average speed is (20+10)/2 = 15 mph. This is incorrect. This would be correct if someone went 20 mph for 10 min then 10 min for 10 min. Let me just do this the long way.

To find the average speed, I need the total distance (got that) and divide by the total time (don't have). I can find the time for the first part of the trip and for the second part. To make this more generic, I am going to call the first speed v₁ and the second speed v₂. I will let the starting position be x = 0 miles. The distance where the bike changes speed will be x₂. The final distance will be x₃. The time for the first part of the trip will be: (note, I will am using v without a vector symbol to denote speed)

Now, to do the same thing for the second part of the trip:

The average speed is now the total distance divided by the total time. The total time is:

The total distance is x₃, so the average speed is:

Before I plug in the numbers - there is on check I can make. Are the units correct? On the top, I have position times v² (units). On the bottom are position and velocity units. These will cancel to just speed units - so that is good. Also, what if v₁= v₂? In this case, the average speed should be v₁. If you plug into the equation above, that is indeed what you get. Now to plug in my numbers.

This is lower than the 15 mph as it should be. The bike was traveling at a lower speed for a longer time than it was at the greater speed. What if I wanted to find the average speed after going v₁ for a time t₁ and then v₂ for a time t₂? This is slightly easier. The total time is just t₁ + t₂. However, I need to find the distances:

Now for the average speed:

This is simply the time-weighted average of the two speeds. Ok - I was afraid of an angry god, so I looked it up. Aeolus is the king of the winds.

rallain Fri, 01/16/2009 - 11:52

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Basics: Relative Velocity

rallain — Wed, 24 Sep 2008 07:19:54 +0000

Basics: Relative Velocity

**pre reqs:** [Vectors and Vector Addition](http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-ad…)

This was sent in as a request. I try to please, so here it is. The topic is something that comes up in introductory physics - although I am not sure why. There are many more important things to worry about. Let me start with an example. Suppose you are on a train that is moving 10 m/s to the right and you throw a ball at 5 m/s to the right. How fast would someone on the ground see this ball? You can likely come up with an answer of 15 m/s - that wasn't so hard right? But let me draw a picture of this situation:

![Screenshot 15](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

The important thing is: If the velocity of the ball is 5 m/s, that is the velocity with respect to what? In the diagram, I listed the velocity of the ball as *v_ball-train* this indicates it is with respect to the train. There are three velocities in this example.

The velocity of the ball with respect to the train
The velocity of the train with respect to the ground
The velocity of the ball with respect to the ground

These three velocities are related by the following:

![relative v](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/relative-…)

**note**: The way I always remember this is to arrange it so that the frames match up on the left side. That is to say v(a-b) + v(b-c) - you can think of this as the "b's" canceling and giving v(a-c).

Clearly this works for the simple case above, but actually it works no matter which direction as long as the equation remains as a **vector** equation. In general, with two reference frames (Say A and B) then you (or I) can say:

![Screenshot 17](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

The most important thing is that these are vectors and must be treated as such. If you treat these vectors as scalars, you will likely get the problem wrong.

Ok. Fine, that makes some sense - but these darn physics problems are killing me (or you). How about an example, everyone loves those. However, if you don't feel comfortable with vectors, go look [at my introduction to vectors](http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-ad…).

**Problem** Suppose you have a boat that travels at a speed of 2 m/s on the water. This boat is to cross a river that is 500 meters wide and has a speed (of the water) of 0.5 m/s. What angle should you aim your boat so that it travels straight across the river (without going downstream at all).

Here is a picture:

![Screenshot 19](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

What is given in the problem? There is the magnitude of the velocity of the boat with respect to the water, the velocity of the water with respect to the ground. There is one other thing that the problem gives. The velocity of the boat with respect to the ground is only in the y-direction. The goal of the problem is to find the angle ? that the boat has to aim. Here is what is given (re-written):

![Screenshot 20](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

Actually, there is one other piece of information that is important. The velocity of the boat with respect to the ground is ONLY in the y direction. I can write this as:

![Screenshot 21](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

Now, let me put these velocities together:

![Screenshot 22](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

Where this is a vector addition equation. To add vectors, I can just add the x-components and then just add the y-components. In this case, I can JUST look at the x-direction:

![Screenshot 23](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

Note that the (m/s) units cancel. Solving for sin(?):

![Screenshot 25](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

So, there you have it. Let me recap what is important:

Start with the relative velocity equation
Write down the velocities you know (as vectors)
Treat the velocities as vectors

See. That is not too bad, is it?

**Final Note:** This is known as Galilean relativity. It works when the velocities of the frames and objects are much less than the velocity of light. (example: a jet going at twice the speed of sound is way slower than light). If the objects are moving close to the speed of light, this stuff does not work.

rallain Wed, 09/24/2008 - 03:19

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thanks for your prompt action.I was teaching relative velocity to someone and I myself was not convinced about my explanations. your exposition makes it crystal clear.

tanks a lot ...........

Thank you so much for your kind information.

With best regards

Sincerely
ashek ullah

Thanks, Me too was looking for a clarification of
idea of relative velocity. I got it here.

your example got some numbers switched. you said water speed of 0.5 m/s. in the equation though you say 1 m/s for Vwater/ground