I Car acceleration if resistance forces don't exist

[jack action]

Where this increase in power come from

Increase in fuel consumption.

They may have a greater capacity to create power, but they don't do magic.

 

[jbriggs444]

Increase in fuel consumption.

No. This is not correct.

The context here is the question of how a rocket can succeed in adding kinetic energy to the payload at a higher rate when going from 100 mph to 200 mph than it can when going from 0 mph to 100 mph.

The fuel consumption is identical for both situations.

The missing bit in the energy accounting is looking at the kinetic energy in the unburnt fuel and in the exhaust stream. For a fast-moving rocket, you have higher kinetic energy in the unburnt fuel and lower kinetic energy in the exhaust stream compared to a slow moving rocket. That's where your extra power comes from. Less (and eventually negative!) power goes to the expended fuel while more power goes to the payload.

If you do the energy accounting properly, you will see that the chemical energy in the expended fuel matches the total increase in kinetic energy of exhaust stream plus rocket. You can further see that this holds regardless of what reference frame you choose to use for the calculation.

 

[PeroK]

No. This is not correct.

The context here is the question of how a rocket can succeed in adding kinetic energy to the payload at a higher rate when going from 100 mph to 200 mph than it can when going from 0 mph to 100 mph.

The fuel consumption is identical for both situations.

... to do the calculations. We have a rocket of mass $M+m$, which expels a small mass $m$ at speed $v$ relative to the rocket. By conservation of momentum, the rocket will accelerate by approximately $\frac m M v$. And the total gain in KE is $$\Delta KE = \frac 1 2 mv^2(1 + \frac m M) \approx \frac 1 2 mv^2$$

If the rocket expels a second mass $m$ with the same relative speed, then we have the same increase in speed and the same gain in KE. In this case, we can see this by changing to the new rest frame of the rocket and observing that we have an identical scenario to the first expulsion. To be precise the acceleration will be slightly greater, as the mass of the rocket has reduced to $M$. As the rocket reduces in mass, these continual mass expulsions will become ever more efficient.

The difference from the previous analysis where the Earth was involved is that we do not have the external large mass to take into account when we change frames.

We can check by doing the analysis in a frame of reference where the rocket has any speed $u$.

Initial momentum $p_0 = (M+m) u$ and initial KE $E_0 = \frac 1 2 (M+m) u^2$.

After the expulsion, we have momentum $p = M(u + \Delta u) - m(v-u)$, hence by conservation of momentum we have (as before) $$\Delta u = \frac m M v$$ And the change in KE is:
$$\Delta E = \frac 1 2 M (u + \Delta u)^2 + \frac 1 2 m(v-u)^2 - \frac 1 2 (M+m) u^2$$
Which after some simplification gives, as before:
$$\Delta E = \frac 1 2 mv^2(1 + \frac m M)$$
Again we see that changing the reference frame does not change the increase in KE.

 

[meekerdb]

So car with rocket engine will gain benefits because of this,in drag racing?
Wait the minute, if rocket engine has constant thurst that mean increase in power with speed?

Where this increase in power come from,how explain this in reference frame of rocket without violate physics laws?

Same answer. Power increases because you're calculating it relative to an increasing velocity. Total energy and momentum are conserved, including the exhaust and chemical energy.

 

[Jurgen M]

Rocket speed is 10km/s,exhaust speed relative to rocket is 10km/s and it is always constant if burn rate is constant.

From initial frame at rest,rocket speed is 10km/s and exhaust speed is zero.
Ke of exhaust is zero.There is no thrust in this frame? confused..

 

[PeroK]

Rocket speed is 10km/s,exhaust speed relative to rocket is 10km/s and it is always constant if burn rate is constant.

From initial frame at rest,rocket speed is 10km/s and exhaust speed is zero.
Ke of exhaust is zero.There is no thrust in this frame? confused..

Thrust is related to momentum conservation.

In your example, the exhaust started at $+10km/s$ and ended at $0 km/s$. In that frame, the expellant lost KE. And the remaining rocket gained KE.

 

[A.T.]

Rocket speed is 10km/s,exhaust speed relative to rocket is 10km/s and it is always constant if burn rate is constant.

From initial frame at rest,rocket speed is 10km/s and exhaust speed is zero.
Ke of exhaust is zero.There is no thrust in this frame? confused..

Thrust is equal but opposite to the rate of change of momentum of the expelled mass, which is non-zero:
https://en.wikipedia.org/wiki/Momentum#Relation_to_force

 

[Jurgen M]

Thrust is related to momentum conservation.

In your example, the exhaust started at $+10km/s$ and ended at $0 km/s$. In that frame, the expellant lost KE. And the remaining rocket gained KE.

Can rocket travel faster then exhaust speed(measured in relation to rocket)?

 

[A.T.]

Can rocket travel faster then exhaust speed(measured in relation to rocket)?

Yes.

 

[PeroK]

Can rocket travel faster then exhaust speed(measured in relation to rocket)?

Speed is frame dependent. In Newtonian mechanics, the speed of the rocket can be any number and the speed of the exhaust be that speed plus or minus the relative speed.

In relativistic mechanics, we have the limitation of the speed of light, but the same applies. The speed of the exhaust may be greater or less than the rocket depending entirely on your chosen frame of reference.

 

[PeroK]

PS if the rocket is less massive than the expellant, then in the initial rest frame of the two the rocket gains more speed.

 

[Jurgen M]

Kinetic energy is frame dependent, just as velocity is.

Momentum is proportional to velocity and is frame dependent too, just as velocity is.

Is this true?

 

[PeroK]

Kinetic energy is frame dependent, just as velocity is.

Momentum is proportional to velocity and is frame dependent too, just as velocity is.

Is this true?

Absolutely. These dynamic quantities must be frame dependent. Acceleration, however, is not frame dependent. That's why we can have Newton's second law (for any inertial frame): $$F = ma$$
In fact, this establishes the important idea that we need to define mass and force in such a way that they are not dependent on the choice of inertial reference frame.

 

[A.T.]

Kinetic energy is frame dependent, just as velocity is.

Momentum is proportional to velocity and is frame dependent too, just as velocity is.

Key is that kinetic energy is non-linearly dependent on velocity, while momentum is linearly dependent on velocity. So changes in kinetic energy are frame dependent, while changes in momentum are frame invariant (for inertial frames).

 

[Jurgen M]

In rocket reference frame there is no increase in power?

 

[PeroK]

In rocket reference frame there is no increase in power?

If we are talking about the power in terms of the energy released by the engine, then that must be invariant.

This thread seems to be going round in circles. You seem to be asking almost random questions without showing much evidence of having digested the content of what is now 66 posts.

 

[Jurgen M]

If we are talking about the power in terms of the energy released by the engine, then that must be invariant.

I need this confirmation.
Generally I see increase in power at rocket as "mathematical manipulation" , I don't see anything useful in real physical sense in it.

 

[A.T.]

I don't see anything useful in real physical sense in it.

You can't see what is useful in real physics without doing real physics.

 

[Jurgen M]

You can't see what is useful in real physics without doing real physics.

If I am understood correctly ,in inertial frame(initial frame at rest) rocket increase power ,in non inertial frame(rocket) power don't increase.

Look to me it is just metter of perspective,I don't know how to call it...

 

[A.T.]

Look to me it is just metter of perspective,I don't know how to call it...

Frame dependent.

 

[Jurgen M]

Problem is that some members at stackechange who comment at special relativity tags, has different comment,answers then here, so it makes even more confusion for me.

 

[A.T.]

Problem is that some members at stackechange who comment at special relativity tags, has different comment,answers then here, so it makes even more confusion for me.

Relativistic mechanics is different, but you asked in the classical physics forum here.

 

[jartsa]

I need this confirmation.
Generally I see increase in power at rocket as "mathematical manipulation" , I don't see anything useful in real physical sense in it.

When an object with mass $m$ moves at speed $v$ it has non-physical "kinetic energy"
$1/2 mv^2$

So a fuel tank with mass million kg and speed million m/s has "kinetic energy" 500000000000000000 J. While its chemical energy is just 10000000000000 J.

Now, when that fuel uses its chemical energy to accelerate itself so that its "kinetic energy" decreases, then the decrease of the "kinetic energy" is a quite large number of Joules.

While the fuel accelerates itself, it must push on something with some force. Now if that something is moving at speed million m/s and is pushed by force million Newtons, then that something is gaining "kinetic energy" at rate 1000000000000 Watts. Million megawatts.

Now where was I going with this? Oh yes, there was the huge decrease of "kinetic energy" of the fuel, and then there was the huge increase of the "kinetic energy" of the thing that the fuel pushed on. It's kind of like whenever there is a huge decrease of energy somewhere, then there must be a huge increase of energy somewhere else.

My point is that the frame dependent "kinetic energy" is quite a lot like real energy.

(To not to be unfair to kinetic energy, the frame independent kinetic energy of a spinning flywheel is perfectly real energy.)

 

[Nugatory]

Problem is that some members at stackechange who comment at special relativity tags, has different comment,answers then here, so it makes even more confusion for me.

It us essential that you understand the classical dynamics before you take on the appreciably hairier relativistic corrections. Otherwise you will never escape the confusion.

 

[cmb]

Let imagine that car with constant 500HP accelerate but resistance forces don't exist (aero drag,internal friction in engine and transmision,tyer rolling resistance etc etc..)
neglect fuel loss over time..

From 0-100km/h take in 4sec and burn 200mL petrol

Will car accelerate from 100-200km/h also in 4sec and burn 200mL petrol?

Here is how I look at it:
Car will also accelerate from 100-200km/h in 4sec and burn 200mL of petrol as well,because ΔV is same in both cases.

But

I know for constant acceleration car need constant Thrust.
I know that car engine produce constant power, Thurst = Power / Velocity, so thrust decrease as speed increase,that implies acceleration will be smaller from 100-200km/h,it will take longer then 4 sec and it will burn more then 200mL of petrol..or same thing from perspective of gears/torques:

Torque at wheel is what accelerate car,we can calculate thurst from wheel torque.
As car increase speed use higher gears which reduce torque at wheels,again thurst is reduced...So what is correct answer and what I am doing wrong?

Constant thrust doesn't equate to constant power.

Work is the product of force times distance. So the further you go, the work done is more for a given amount of force (thrust).

You defined your engine as a fixed power, so as it accelerates, the thrust will decrease even if the power remains constant, simply because power is thus the product of force times velocity.

If you car was powered by a constant thrust jet engine, you'd have been right.

 

[cjl]

Rocket speed is 10km/s,exhaust speed relative to rocket is 10km/s and it is always constant if burn rate is constant.

From initial frame at rest,rocket speed is 10km/s and exhaust speed is zero.
Ke of exhaust is zero.There is no thrust in this frame? confused..

There is thrust. The exhaust has zero energy, but when it was fuel traveling with the rocket, it had a very large amount of energy (since it was traveling 10km/s). Therefore, when the rocket engine exhausted the fuel, it removed 100% of the fuel's energy, and the rocket gained all the energy that was previously in the fuel. Since the rocket is lighter now, but still has just as much energy, it must be traveling faster.

 

[sophiecentaur]

If you car was powered by a constant thrust jet engine, you'd have been right.

With constant thrust, the constant rate of Momentum Increase would be the relevant quantity. But even that will fail at high speeds over the ground.

 

[Jurgen M]

now case with 2 constanat speeds, 100km/h and 200km/h

If aero drag don't exist, only exist tire resistance which is constant(lets say 50N) for all speeds, car will again use more fuel when driving at constant speed at 200km/h then on 100km/h?
P=F x v

constant 100km/h
P= 50N x 27m/s = 1350W

constant 200km/h
P=50N x 55m/s = 2750W

Again, it is unintutitve how fuel consumptin increase if drag force(50N) not changed.. That mean you use more fuel/power just becuase you travel faster even if resistance force don't change at all..

Am I right?

 

[jbriggs444]

now case with 2 constanat speeds, 100km/h and 200km/h

If aero drag don't exist, only exist tire resistance which is constant(lets say 50N) for all speeds, car will again use more fuel when driving at constant speed at 200km/h then on 100km/h?

More fuel per unit distance covered? No.
More fuel per unit time taken? Yes.

 

[Jurgen M]

More fuel per unit distance covered? No.
More fuel per unit time taken? Yes.

Yes now make sense.Back to orignal acceleration question.
in acceleration from 100 to 200km/h car use more fuel per unit distance coverd than from 0 to 100km/h?
(remeber car display consumption in L/100km)
(if aero drag don't exist , just constant tire drag exist that not change with speed)

 

[jbriggs444]

Back to orignal acceleration question.
in acceleration from 100 to 200km/h car use more fuel per unit distance coverd than from 0 to 100km/h?

That is going to depend on the rate of acceleration. We are assuming, per your stipulation, that there is negligible air resistance.

If the acceleration rate is held constant then the acceleration just amounts to an additional fixed force. The change in current speed will not have any effect on fuel consumption per unit distance.

If the acceleration rate is allowed to decline (perhaps engine power is limited) then the acceleration would amount to a large force at low speed and a smaller force at high speed. You would be more fuel efficient per unit distance at higher speed because of this.

We are assuming, perhaps counter-factually, that the engine has constant efficiency regardless of power setting. [My understanding is that real engines usually have a sweet spot where power is obtained most efficiently. Higher power output than that comes at the expense of increased fuel used per unit of energy released. Lower power output than that is also less efficient because the engine is wasting much of its output on the power required to just keep running]

You might concern yourself with the fact that kinetic energy goes with the square of velocity and it seems like you are burning 1 unit of energy to get to 100 km/h and 3 additional units of energy to get to 200 km/h. Yes indeed. But (for constant acceleration), you cover $\frac{1}{2}at^2$ of ground getting to 100 km/h and an additional $\frac{3}{2}at^2$ getting from 100 km/h to 200 km/h. That is a factor of three to one in distance which nicely matches the factor of three to one in kinetic energy.

 

[sophiecentaur]

Kinetic energy raise with speed for rocket as well,but rocket(constant thrust) will have same acceleration and fuel burn from 0-100km/h and from 100-200km/h, so know that Ke raise with speed don't tell me nothing.
If we neglect fuel loss over time.

The most difference between car and rocket from what I see is that rocket has constant thrust and car has constant power(but reduced thrust with speed)..
reduced thrust implies reduced acceleration..

In my opinion, you have tried to leap in at the deep end of the topic. Cars, rockets and even electric motors are not ideal for starting on this problem.
If you want to make your car / mass move under constant acceleration then you need an ideal model in the form of (the equivalent of) a force meter and a speed controlled tractor of some sort. You can write correct equations for that.
For “constant power” you can adjust the speed of the tractor to give an appropriate force. But that’s not an arm waving process.
Also, for these elementary models, you need slow speeds (a very few m/s) so you can at least omit air drag.

 

[Jurgen M]

That is going to depend on the rate of acceleration. We are assuming, per your stipulation, that there is negligible air resistance.

If the acceleration rate is held constant then the acceleration just amounts to an additional fixed force. The change in current speed will not have any effect on fuel consumption per unit distance.

If the acceleration rate is allowed to decline (perhaps engine power is limited) then the acceleration would amount to a large force at low speed and a smaller force at high speed. You would be more fuel efficient per unit distance at higher speed because of this.

We are assuming, perhaps counter-factually, that the engine has constant efficiency regardless of power setting. [My understanding is that real engines usually have a sweet spot where power is obtained most efficiently. Higher power output than that comes at the expense of increased fuel used per unit of energy released. Lower power output than that is also less efficient because the engine is wasting much of its output on the power required to just keep running]

You might concern yourself with the fact that kinetic energy goes with the square of velocity and it seems like you are burning 1 unit of energy to get to 100 km/h and 3 additional units of energy to get to 200 km/h. Yes indeed. But (for constant acceleration), you cover $\frac{1}{2}at^2$ of ground getting to 100 km/h and an additional $\frac{3}{2}at^2$ getting from 100 km/h to 200 km/h. That is a factor of three to one in distance which nicely matches the factor of three to one in kinetic energy

It seems all of you forget in this topic that car display consumtion in L/100km( fuel per unit of coverd distance) or maybe I need to note this.

We can set gear ratio that car allways work with full throttle from 6000rpm to 8000rpm, for both runs, 0-100 and 100 -200..

My two inital questions allways was
1.If someone ask me what is the main reason why every car accelerate significialy slower from 100-200 then from 0-100 if in both case deliver same power.
(we state here that car will behave the same even we neglect air,so aero drag is no reason,plus I knew that aero drag is not such a big to slow down acceleration so much,air is not water)
2. will car computer display higher L/100km if we accelarate from 100-200 then 0-100.
again with no aero drag

 

[jbriggs444]

1.If someone ask me what is the main reason why every car accelerate significialy slower from 100-200 then from 0-100 if in both case deliver same power.
(we state here that car will behave the same even we neglect air,so aero drag is no reason,plus I knew that aero drag is not such a big to slow down acceleration so much,air is not water)

2. will car computer display higher L/100km if we accelarate from 100-200 then 0-100.
again with no aero drag

OK. So the assumption of constant acceleration is out the window. We have constant power instead.

I would answer 1 above by noting that for constant power, the work done per unit time is constant. The amount of work done per unit time is force times the distance covered in that unit time. That is to say that power $P = \vec{F} \cdot \vec{v}$.

Increase velocity $\vec{v}$ and hold power $P$ constant and you'll need to reduce force $\vec{F}$.

Acceleration $\vec{a}$ is given by $\frac{\vec{F}}{m}$. Reduce propulsive force $\vec{F}$ and you've reduced forward acceleration $\vec{a}$.

Or, if you prefer, you had to shift to a higher gear. Less available force at the wheels. The faster you go, the less acceleration results.

I would answer 2 by noting that your scenario is extraordinarily simple. The engine runs at constant power. Fuel consumption per unit time is fixed. You get the best fuel consumption per unit distance when you cover the greatest possible distance in that time. It is a direct proportionality. Fuel efficiency (per unit distance) is directly proportional to speed in this situation.

This is entirely independent of aerodynamic drag. Lots of drag, little drag. It is all the same. As long as you are running at a given constant power, all that matters for fuel efficiency is how fast you are going. The faster the better. Aerodynamic drag only enters in if it limits how fast you can go. Or if it forces you to use a bigger engine with a greater constant power output.

 

[Jurgen M]

Also, for these elementary models, you need slow speeds (a very few m/s) so you can at least omit air drag.

I can set my question in wacuum room.I.C. engine has gerabox which has gear ratios that keep engine full thortlle working allways between 6000rpm-8000rpm when lift weight.

Everthying is set in wacuum room,aero drag don't exist, only 50N of weight and some small constant drag friction force from rope,gearbox and pulley, that don't change with speed..Question will be

1. Why weight acceleration is smaller when weight lift up from 100-200km/h then from 0-100km/h if drag force and weight is unchanged?2.Is L/100km (fuel per unit of distance of weight) same from 0-100 vs 100-200km/h

1.
For what I can see, tension in rope represent thrust for weight.
Logicaly tension in rope will be higher in 1gear then in 6gear,so engine reduce thrust as gears are changed,this I see as reason why weight reduced acceleration when lifitng up..

If rocket with constat thrust pull rope ,acceleration will be the same.

2.
L/100km will be smaller from 100-200km/h

 

[jbriggs444]

I can set my question in wacuum room.

1. Why weight acceleration is smaller when weight lift up from 100-200km/h then from 0-100km/h if drag force and weight is unchanged?

The diagram you show has a constant force device. However, you apparently want to use some gears or pulleys (not shown) so that the "50 N" thrust in first gear becomes, perhaps, a 10 N thrust in fifth gear.

This would not be a completely accurate representation of a constant power device, but adequately close. The rate of descent of the weight would change within each fixed gear ratio. As a result, the power would decrease with each up-shift and then increase until the next up-shift. One would change gears/pulleys to maintain a reasonably constant descent rate for the weight.

1.
For what I can see, tension in rope represent thrust for weight.
Logicaly tension in rope will be higher in 1gear then in 6gear,so engine reduce thrust as gears are changed,this I see as reason why weight reduced acceleration when lifitng up..

Yes, that explanation works.

If rocket with constat thrust pull rope ,acceleration will be the same.

Let us not talk about rockets until we have wheeled propulsion squared away. Rockets conserve energy too. It is just that you have to look for the energy in different buckets.

2.
L/100km will be smaller from 100-200km/h

Yes.

 

[Jurgen M]

OK. So the assumption of constant acceleration is out the window. We have constant power instead.

I would answer 1 above by noting that for constant power, the work done per unit time is constant. The amount of work done per unit time is force times the distance covered in that unit time. That is to say that power $P = \vec{F} \cdot \vec{v}$.

Increase velocity $\vec{v}$ and hold power $P$ constant and you'll need to reduce force $\vec{F}$.

Acceleration $\vec{a}$ is given by $\frac{\vec{F}}{m}$. Reduce propulsive force $\vec{F}$ and you've reduced forward acceleration $\vec{a}$.

Or, if you prefer, you had to shift to a higher gear. Less available force at the wheels. The faster you go, the less acceleration results.

I would answer 2 by noting that your scenario is extraordinarily simple. The engine runs at constant power. Fuel consumption per unit time is fixed. You get the best fuel consumption per unit distance when you cover the greatest possible distance in that time. It is a direct proportionality. Fuel efficiency (per unit distance) is directly proportional to speed in this situation.

This is entirely independent of aerodynamic drag. Lots of drag, little drag. It is all the same. As long as you are running at a given constant power, all that matters for fuel efficiency is how fast you are going. The faster the better. Aerodynamic drag only enters in if it limits how fast you can go. Or if it forces you to use a bigger engine with a greater constant power output.

Now I agree with everything, wish I before insist to use only fuel per units of distance,like cars allways show..

Then rocket enigne is better for drag racing, don't loose thrust with speed

 

[jbriggs444]

Then rocket enigne is better for drag racing, don't loose thrust with speed

Not really. It is hard to do an apples to apples comparison when comparing wheeled propulsion to rocket propulsion.

With high velocity rockets propelling low velocity vehicles you are wasting most of your fuel energy into the exhaust stream. Your fuel efficiency is atrocious. If you could use all of the raw energy in the rocket's exhaust to push backward on the ground rather than backward on the exhaust stream you would do dramatically better. At least up until the vehicle speed is up to a significant fraction of the exhaust velocity -- well past the speed of sound.

Also, a drag race benefits most from high acceleration early. High acceleration late will not help as much.

Normally one judges drag races on elapsed time, not fuel consumption.

 

[Jurgen M]

The diagram you show has a constant force device.

You mean constant power ?

This would not be a completely accurate representation of a constant power device, but adequately close.

I.C. engine produce constant power only if you keep it at same rpm, but I assume that from 6000-8000rpm has aprox constant power...

 

[jbriggs444]

You mean constant power ?

You showed a 50 N weight on a string pulling a cart. That is a constant force device. It is not a constant power device.

The faster a weight falls, the faster gravitational potential energy is released and the more power is transmitted by tension in the more rapidly moving string. $P = \vec{F} \cdot \vec{v}$.

I.C. engine produce constant power only if you keep it at same rpm, but I assume that from 6000-8000rpm has aprox constant power...

If you are running your engine at max power (and if max power is not at red line), then the rate of change of power with respect to RPM must be zero there. So yes, if 7000 RPM is max power then 6000 to 8000 might be approximately max power.

 

[Jurgen M]

You showed a 50 N weight on a string pulling a cart. That is a constant force device. It is not a constant power device.

No this not string pulling cart.
Engine/gearbox stay at place and winding wheel lift up weight with rope like cran that has I.C. engine for lifting.Weight is accelerate up.

 

[jbriggs444]

No this not string pulling cart.
Engine/gearbox stay at place and winding wheel lift up weight with rope like cran that has I.C. engine for lifting.Weight is accelerate up.

If you have constant 50 N force, it is a constant force device. Period. End Of Story.

Maybe if it was a 5 kg mass, it could be constant power. But it's not 5 kg. It's labeled as 50 N.

 

[Jurgen M]

If you have constant 50 N force, it is a constant force device. Period. End Of Story.

Maybe if it was a 5 kg mass, it could be constant power. But it's not 5 kg. It's labeled as 50 N.

It has mass of 5kg, on the Earth this aprox 50N, and when accelerate up tension in rope is bigger then 50N...

 

[Jurgen M]

I suspect that he doesn't ride a bicycle!

Indeed bike can travel very fast if reduce aero drag, 183mph (295km/h) !

 

[bhobba]

A car can't have ANY acceleration if there is no resistance anywhere including the tire/road resistance because the wheels would just spin and the car would go nowhere.

What was it Meatloaf said - you took the words right out of my mouth (the rest doesn't apply here ). Seriously the only reason a car can move is the resistance between the tyres and the road. If there is none all the wheels would do is spin - as you mentioned. I seem to recall Feynman discussing it somewhere in his lectures, which of course anyone interested in applied math/physics should read (but not after a usual course in physics without going into why).

Thanks
Bill

 

[bhobba]

Problem is that some members at stackechange who comment at special relativity tags, has different comment,answers then here, so it makes even more confusion for me.

With my mentor's hat on first, this is a thread about classical Newtonian mechanics; relativistic considerations are not appropriate. But different answers to the same question can be confusing; so just a passing comment that there are different approaches to relativity that can lead to different ways of explaining things. I think these days the explanation based on the symmetries of an inertial frame gets to the heart of the matter best:
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

If you would like to discuss the matter further please start a thread in the relativity forum.

Thanks
Bill