A problem regarding Galilean relativity

[CherryWine]

Homework Statement

Two bodies are moving on the same line. When they move away from each other the distance between them changes for 16m in a time interval of 3 s (Δd₁ = 16 m ; Δt₁ = 3 s). When they move towards each other the distance between them changes for 3 m in a time interval of 3 s (Δd₂ = 3 m ; Δt₂ = 3 s). What are the respective speeds of the two bodies?

Homework Equations

Basic kinematic and Galilean relativity equations.

The Attempt at a Solution

So, I've reasoned that if the speeds of the bodies are constant, then in both cases, whether they are moving towards or away each other, the relative speed should be the same, because in both cases it is given by V_relative=V₁+V₂, since their respective velocity vectors point in different directions in both cases. From that it follows that if the relative speeds are the same in both cases, then this equality holds Δd₁/ Δt₁ = Δd₂/Δt₂ , but it obviously does not when the numbers given in the problem are plugged in.

The solution in the book assumes that the relative speeds are not the same in the two cases, saying that in the case when they move away from each other the relative speed is given by V_relative=V₁+V₂ but however in the case when they move towards each other the relative speed is given by V_relative=V₁-V₂ which I don't understand why.

 

[Hamal_Arietis]

It is wrong! Because the the motion of each body has acceleration. veclociy increases or decreases diferrence so that v_relative increases or descreses too. Your case just right when acceleration=0

 

[CherryWine]

It is wrong! Because the the motion of each body has acceleration. veclociy increases or decreases diferrence so that v_relative increases or descreses too. Your case just right when acceleration=0

What ?! The mentioned speeds/velocities are constant, meaning the acceleration is zero.

 

[Hamal_Arietis]

ahh sorry :) first they move to near. and after they catch they move to far. and v =v1+v2 in 2 cases

 

[CherryWine]

ahh sorry :) first they move to near. and after they catch they move to far. and v =v1+v2 in 2 cases

They do not "catch" in any case. Imagine that their initial distance is infinite. Mentioned scenarios are separate.

 

[Hamal_Arietis]

Sorry my English isn't good so i can't read this problem correctly. After used gg translation. I answer you that: When they move away the relative velocity v=v1+v2. End when they move towards v=v1-v2.

 

[Hamal_Arietis]

Did you studied about the vector? It is total vector formula. But if you didnt. You can see this solution:Suppose that v1>v2
Move away:the distance they move in denta t
s1=v1denta t; s2=v2 denta t=> denta d=s1+s2=(v1+v2)denta t
Move towards: denta d=s1-s2=(v1-v2) denta t
v1+v2 and v1-v2 are relative velocity in each case.

 

[CherryWine]

Did you studied about the vector? It is total vector formula. But if you didnt. You can see this solution:Suppose that v1>v2
Move away:the distance they move in denta t
s1=v1denta t; s2=v2 denta t=> denta d=s1+s2=(v1+v2)denta t
Move towards: denta d=s1-s2=(v1-v2) denta t
v1+v2 and v1-v2 are relative velocity in each case.

Can you please explain why would Δd = s₁-s₂ in the case of them moving towards? Suppose they move with the same velocity, thus they cover equal distances in equal time intervals, therefore by your reasoning Δd would be zero, when obviously it is not!

 

[Hamal_Arietis]

I think move towards means same direction. Two vector v1,v2 are the same direction.
if v1=v2 mean denta d=0 => the distance between two bodies dont
change. But in this problem v1 and v2 are different.

 

[CherryWine]

I think move towards means same direction. To vector v1,v2 are the same direction.
if v1=v2 mean denta d=0 => the distance between two bodies dont
change. But in this problem v1 and v2 are different.

The problem is only solvable if the "move towards" case is considered as both bodies moving in the same direction (only one of them facing the other). However, the problem arose since this is not clear from the text of the problem.