Determining the interval between two F1 cars

[ilasus]

Two cars A, B move in the same direction with the constant speeds v, u (v < u) on a rectilinear OM circuit, specially built for Formula 1 cars. Assuming that at the initial moment (t = 0), the two cars are on the starting line O (x = 0) of the OM circuit and that, during t, the cars A, B travel the distances

x₁ = v t, x = u t

in relation to O, the question is what is the interval time t_AB between machines A and B.

My solution. In the case of Formula 1 cars, the time between them is measured using transponders that transmit radio waves to timing loops placed on the circuit: when a car passes over a timing loop, a system will begin to time how long until it takes as the next car to arrive exact the same spot, when the time between the two cars is established - which can be constant or variable over a certain distance. In the case of machines A, B, the time t_AB between them is uniformly increasing and is measured from the moment t₁ when machine B is at a distance x₁ from the starting line O, until moment t when machine A is at the same distance x₁ from O, so the time t_AB is given by the relation

t_AB = t – t₁

where

t₁ = x₁/u = (vt)/u = (v/u²) x

What do you think?

 

[Delta2]

In my opinion the time interval $t_{AB}$ is equal to the time it takes for the second machine (the machine with the smaller speed) to cover their difference in distance, that is $$t_{AB}=\frac{x-x_1}{v}=\frac{x}{v}-\frac{x_1}{v}=\frac{u}{v}t-t=t(\frac{u}{v}-1)$$

 

[nasu]

x₁ = v t, x = u t

in relation to O, the question is what is the interval time t_AB between machines A and B.

First you have to define what is the meaning of the "interval time between machines A and B". Without this, anything goes. It is your personal definition.

 

[ilasus]

The ”time interval t_AB” simply means the time gap btween machines A and B – it is noted on the side of the live leaderboard on TV during a race F1. But how is it correct calculated, like me, or like Delta2? Thank you.

 

[nasu]

You will have to ask the people who use this "quantity". There is no meaningful discussion about a quantity whose definition is unknown.

 

[Delta2]

You will have to ask the people who use this "quantity". There is no meaningful discussion about a quantity whose definition is unknown.

It isn't unknown definition, he gave his definition , I gave mine, maybe we can prove that the two are equivalent.

Now that I read post#1 more carefully, the definition there is exactly what they do in F1, so I think it is correct.

Don't know how i came up with my own definition lol.

 

[nasu]

Then what is the point to ask how is it calculated, if the definition is really given?

 

[berkeman]

It isn't unknown definition, he gave his definition , I gave mine, maybe we can prove that the two are equivalent.

I think they are equivalent. The lap times (and finishing positions) are determined by when the front of the vehicle or person crosses the finish line (since the front of the vehicles or persons are staged right at the start line to begin the race). That is true in pretty much all kinds of racing that I'm familar with.

https://www.westend61.de/en/imageView/CAIF18490/runners-crossing-finish-line-on-track

 

[haruspex]

I see no definition of $x_1$. I am guessing it is the length of the circuit.
Whatever, we want the time difference in covering distance $x_1$, but in terms of what?
Seems to me "x" and "t" are not 'given' values, they are just variables used to describe general relationships. So the form of the answer cannot include either of those. This leaves only u, v and $x_1$.

 

[ilasus]

For example, suppoze that u = 5 m/s, v = 4 m/s, t = 10 s, x₁ = 40 m and x = 50 m (x = 50 m is length of the circuit). In this case, I consider that t_AB = (x - x₁)/u = 2 s, and Delta2 says that t_AB = (x - x₁)/v = 2.5 s. So our views are not compatible. How is it correct?

 

[haruspex]

For example, suppoze that u = 5 m/s, v = 4 m/s, t = 10 s, x₁ = 40 m and x = 50 m (x = 50 m is length of the circuit). In this case, I consider that t_AB = (x - x₁)/u = 2 s, and Delta2 says that t_AB = (x - x₁)/v = 2.5 s. So our views are not compatible. How is it correct?

As I wrote in post #9, "x" cannot feature in the answer. It has no specific value. It is only used as the generic position of the faster car at an arbitrary time t.
As implied in the problem statement, the apparatus detects the time at which the faster car reaches a particular distance $x_1$, then measures how long it takes the slower car to reach the same point. Express that time in terms of $x_1, u, v$.

 

[Delta2]

the apparatus detects the time at which the faster car reaches a particular distance x1, then measures how long it takes the slower car to reach the same point

Yes I think that's exactly what they do in F1 and other racing championships.

 

[ilasus]

And how is it correct: t_AB = (x-x₁)/v or t_AB = (x-x₁)/u ?

 

[Delta2]

if I follow @haruspex advice at post #11, I think the correct answer is $\frac{x_1}{v}-\frac{x_1}{u}$. x should not feature in the answer

 

[haruspex]

And how is it correct: t_AB = (x-x₁)/v or t_AB = (x-x₁)/u ?

As I explained in post #9 and in post #11, neither of those is a correct answer.

 

[haruspex]

if I follow @haruspex advice at post #11, I think the correct answer is $\frac{x_1}{v}-\frac{x_1}{u}$. x should not feature in the answer

I wanted to give the OP a bit more opportunity get there.

 

[ilasus]

Haruspex:
- I don't understand your answer # 9 and # 11
- What does "OP" mean?
- "get there" - where there?
- Can you give a simple and clear example to understand what you mean?

 

[haruspex]

Haruspex:
- I don't understand your answer # 9 and # 11
- What does "OP" mean?
- "get there" - where there?
- Can you give a simple and clear example to understand what you mean?

"OP", on these forums, means either Original Post (post #1 in the thread) or Original Poster, the person who created the thread: you in this case.
Getting there means arriving at the answer.

The way the problem is stated in post #1 is rather confusing. I'll attempt to rewrite it to make it clearer:

"Two cars A, B move in the same direction with the constant speeds v, u (v < u) on a rectilinear OM circuit, specially built for Formula 1 cars. Assume that at t = 0 the two cars are on the starting line, O.
So at any given later time t, car A has gone distance vt and car B has gone distance ut.

... the time t_AB between them ... is measured from the moment ... when machine B is at a distance x₁ from the starting line O, until moment ... when machine A is at the same distance x₁ from O"

Note that "x" no longer appears and that "t" is not a specific time. The only data you have to work with are u, v and x₁. You need to find an expression for t_AB which only uses those three variables.

 

[ilasus]

So at any given later time t, car A has gone distance vt and car B has gone distance ut.

So at any given later time t, car A has gone distance x₁=vt and car B has gone distance x=ut. Therefore, if v=au, (0<a<1), then x₁=ax. Specifically, if u=5m/s, v=4m/s, then a=4/5 and x₁=(4/5)x. So:
- if t = 1s, then x₁ = 4m, x = 5m, t₁ = (x-x₁) / u = 1/5 seconds
- if t = 2s, then x₁ = 8m, x = 10m, t₁ = (x-x₁) / u = 2/5 seconds
.....
- if t = 10s, then x₁ = 40m, x = 50m, t₁ = (x-x₁) / u = 10/5 seconds. Understand?

 

[ilasus]

What you think about this video ?

 

[haruspex]

What you think about this video ?

Is it relevant to the question in post #1? If so, what is it about the video that you seek my opinion on?

If you are still interested in the answer to post #1, please post your attempt using my hints in posts #11 and #18.

 

[ilasus]

Is it relevant to the question in post #1? If so, what is it about the video that you seek my opinion on?

Yes.

Car A in post #1 is O' in the video
Car B in post #1 is M in the video
And the time t_AB from post #1 is t₂ in the video.

In the video I also answer the question from post #1.

But you can only refer to the video.
So what do you think about the content of the video?

 

[haruspex]

Yes.

Car A in post #1 is O' in the video
Car B in post #1 is M in the video
And the time t_AB from post #1 is t₂ in the video.

In the video I also answer the question from post #1.

But you can only refer to the video.
So what do you think about the content of the video?

I don’t see the relevance. F1 cars do not move at relativistic speeds.

 

[ilasus]

Thanks for the reply, but the video doesn't talk about relativistic velocities or TR. In order to see the relevance you are referring to and to present your own opinion, you should watch the video first. If you don't like videos, then read the attached PDF article.

 

[jbriggs444]

The art of solving physics homework problems is the art of slicing away the irrelevant and concentrating on the relevant details. Reducing the problem to one (or a few) simple equations which can be solved for the desired answer.

A video expounding on the Galilean (Newtonian mechanics) and Lorentz (special relativity) transforms just adds unnecessary complexity. One uses those transforms when shifting from one frame of reference to another. This problem can be solved from a single frame of reference: the track frame.

What is needed is simplicity. Which @haruspex is laboring to provide by concentrating on a clear formulation of the problem. "A question well asked is half solved."

 

[ilasus]

What is needed is simplicity.

The reality is not quite as simple as we would like.

 

[jbriggs444]

The reality is not quite as simple as we would like.

But the reality of physics homework questions is.

 

[haruspex]

In the video I also answer the question from post #1.

I will not watch a video just to find your answer to the question in post #1. That answer is a few lines of algebra.

 

[ilasus]

OK

 

[ilasus]

In the video I refer to a mobile M moving with speed u on a road R and I state that the movement with speed u of mobile M on the platform R' is virtual. Hence the formulas (4), (4') which, for u=c, are identified with the Lorentz transformations. So in my video it is a deduction by methods of classical physics, of Lorentz transformations.