I Car acceleration if resistance forces don't exist

[Jurgen M]

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?

 

[Lnewqban]

Welcome!
A real engine is a complicated example, because it can't deliver work in a linear way or for a broad range.
Perhaps, using an electrical motor will give you better results.

 

[Jurgen M]

Welcome!
A real engine is a complicated example, because it can't deliver work in a linear way or for a broad range.
Perhaps, using an electrical motor will give you better results.

Lets assume engine produce constant power all the time,to make things easier

 

[phystro]

You are correct, since dv is the same, acceleration will also be the same, and the force F=ma will also be the same. You could even say that it might be even easier since the weight of the car will reduce because of the fuel that is lost.

What you are doing wrong is this:
"I know that car engine produce constant power, Thurst = Power / Velocity, so thrust decrease as speed increase"

here you should use dv instead of v and you will get the same result.

 

[phinds]

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.

 

[Jurgen M]

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.

Yes I know that,but let assume tyer rolling resistance is small so we can neglect it..
Point of question is important.

 

[Jurgen M]

You are correct, since dv is the same, acceleration will also be the same, and the force F=ma will also be the same. You could even say that it might be even easier since the weight of the car will reduce because of the fuel that is lost.

So you want to say the only reason why car accelerate significantly slower from 100-200km/h then 0-100km/h in real life are resistance forces(aero drag,tyer friction etc)?

So this first two answers on this site below are wrong?
https://physics.stackexchange.com/q...el-with-equal-acceleration-Δv-but-at-differen

 

[phinds]

Point of question is important.

Accurate statement of the problem you are actually trying to solve is ALSO important.

 

[PeroK]

Yes I know that,but let assume tyer rolling resistance is small so we can neglect it..
Point of question is important.

You do not get a constant acceleration from constant power in the case of tyres on a road. You can see this in a number of ways:

First, KE is $\frac 12 mv^2$. As the speed increases, so more energy is needed for each increment. E.g. three times as much energy is needed to accelerate from $1m/s$ to $2m/s$ as from rest to $1m/s$.

Second, in this case $P = Fv$ and we see that the accelerating force decreases with speed for constant power.

Third, if you analyse the car-Earth system, you see further justification for this.

This is not true of all propulsion mechanisms - e.g. throwing stuff out of the back of the car gives the same acceleration regardless of the speed relative to the Earth.

 

[PeroK]

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.

Acceleration and rolling through static friction can be very efficient. Hence the importance of the wheel!

And, indeed, tyres that grip well and have a high static friction allow greater acceleration.

 

[Jurgen M]

You do not get a constant acceleration from constant power in the case of tyres on a road. You can see this in a number of ways:

First, KE is $\frac 12 mv^2$. As the speed increases, so more energy is needed for each increment. E.g. three times as much energy is needed to accelerate from $1m/s$ to $2m/s$ as from rest to $1m/s$.

This is case where reference frame is road,but that frame is irrelevant for this case because this is not place where acceleration is physicaly applyed.
Energy is frame dependet.

From car perspective, car speed is zero even it stay or travel at 100km/h.It only appears that kinetic energy rise from the "rest frame" from which you initially started accelerating but that is not physically relevant because that is not where you are going to be physically applying the acceleration.

Isnt it?

 

[PeroK]

This is case where reference frame is road,but that frame is irrelevant for this case because this is not place where acceleration is physicaly applyed.
Energy is frame dependet.

From car perspective, car speed is zero even it stay or travel at 100km/h.It only appears that kinetic energy rise from the "rest frame" from which you initially started accelerating but that is not physically relevant because that is not where you are going to be physically applying the acceleration.

Isnt it?

It is physically relevant because when you change to the rest frame of a moving car, so the Earth is rotating faster depending on the speed of the car. And, as the acceleration mechanism involves pushing on the Earth, so the energy required to accelerate the Earth by the next increment increases.

That's why I included the point about analysis in the Earth-car frame.

 

[jbriggs444]

@PeroK has this one right. Cars work by pushing on the road.

As you go faster, it takes more power to exert a constant rearward thrust against the road.

You either run the engine faster, slurping up more air and burning more fuel or you shift into a higher gear, reducing your torque at the wheels and thereby taking longer and burning more fuel that way.

 

[Jurgen M]

@PeroK has this one right. Cars work by pushing on the road.

As you go faster, it takes more power to exert a constant rearward thrust against the road.

You either run the engine faster, slurping up more air and burning more fuel or you shift into a higher gear, reducing your torque at the wheels and thereby taking longer and burning more fuel that way.

But why would object will be harder to accelerate when object travel at higher speeds if speed is relative term?
I don't understand this concept at all..

So you don't agree with comments from member @m4r35n357 in site below?
https://physics.stackexchange.com/q...ct-at-higher-speeds-if-speed-is-relative-term

 

[jbriggs444]

But why would object will be harder to accelerate when object travel at higher speeds if speed is relative term?
I don't understand this concept at all..

Speed relative to the road is absolute. It does not depend on reference frame.

The energy required to go from 100 mph relative to the road to 200 mph relative to the road can be calculated in any reference frame you like. The answer will be the same in all of them. It will be "invariant".

Edit to add...

As has been pointed out, if you use rocket propulsion, the road becomes irrelevant as a reference. It takes the same fraction of remaining gross vehicle weight to accelerate from zero to 100 mph as it cakes to accelerate from 100 mph to 200 mph using rocket propulsion.

But with rocket propulsion, you need a fuel load that scales exponentially with desired total delta v. If you want a very high ratio of craft velocity to exhaust velocity, the mathematics works out poorly for you.

 

[PeroK]

So you don't agree with comments from member @m4r35n357 in site below?
https://physics.stackexchange.com/q...ct-at-higher-speeds-if-speed-is-relative-term

I suspect that he doesn't ride a bicycle!

It's a common misconception. It does, however, apply to a photon rocket in space, for example, but not a car on an Earthbound road.

 

[Jurgen M]

Speed relative to the road is absolute. It does not depend on reference frame.

The energy required to go from 100 mph relative to the road to 200 mph relative to the road can be calculated in any reference frame you like. The answer will be the same in all of them. It will be "invariant".

Edit to add...

As has been pointed out, if you use rocket propulsion, the road becomes irrelevant as a reference. It takes the same fraction of remaining gross vehicle weight to accelerate from zero to 100 mph as it cakes to accelerate from 100 mph to 200 mph using rocket propulsion.

But with rocket propulsion, you need a fuel load that scales exponentially with desired total delta v. If you want a very high ratio of craft velocity to exhaust velocity, the mathematics works out poorly for you.

So the main problem is that engine produce constant power not constant thrust?

In real life car accelerate from 100-200km/h significally slower then from 0-100km/h, is the main contribution for this ,resistance forces (aero drag and tyer friction) or Ke=1/2 m v "2

How rocket "avoid" Ke=1/2 m v" 2 rule?

 

[PeroK]

In real life car accelerate from 100-200km/h significally slower then from 0-100km/h, is the main contribution for this ,resistance forces (aero drag and tyer friction) or Ke=1/2 m v "2

For an efficient, streamlined vehicle it's mostly the effect of relative speed with the road. You can see that by looking at your ability to freewheel at high speed even though you cannot accelerate further. If you take your foot off the gas, you can see the resistance forces are not massive.

How rocket "avoid" Ke=1/2 m v" 2 rule?

it doesn't avoid the rule. When you take the loss on KE of the expellant into account it all balances. Whatever reference frame you use.

That's why the propulsion system matters.

I'd encourage you to do the calculations for both methods. In neither case treating the car or rocket as a closed system. In one case, you have the Earth. And in the other the expellant.

 

[Jurgen M]

For an efficient, streamlined vehicle it's mostly the effect of relative speed with the road. You can see that by looking at your ability to freewheel at high speed even though you cannot accelerate further.

Can you explain this effect of relative speed with road?
Why you can't accelerate further even if you freewheel?

 

[PeroK]

Can you explain this effect of relative speed with road?
Why you can't accelerate further even if you freewheel?

You need to do the calculations. You can postulate zero resistance, zero energy loss, but only a very small further acceleration for a given power. If you use the rest frame of the car, then the Earth has KE $\frac 1 2 Mv^2$ and that is hard to change if you want more speed relative to the road.

Note that conservation of momentum of the Earth-car system is key.

There comes a point where physics cannot be explained any further and you need to do the calculations for yourself.

 

[Jurgen M]

You need to do the calculations. You can postulate zero resistance, zero energy loss, but only a very small further acceleration for a given power. If you use the rest frame of the car, then the Earth has KE $\frac 1 2 Mv^2$ and that is hard to change if you want more speed relative to the road.

Note that conservation of momentum of the Earth-car system is key.

There comes a point where physics cannot be explained any further and you need to do the calculations for yourself.

Dont understand you,but never mind..
This topic is too complicated for my logic,problem is that I still can't understand why object is harder to accelerate if travel with higher speed.
In my brain any constant speed and zero speed is same thing,from car perspective/frame

 

[PeroK]

Dont understand you,but never mind..
This topic is too complicated for my logic,problem is that I still can't understand why object is harder to accelerate if travel with higher speed.
In my brain any constant speed and zero speed is same thing,from car perspective/frame

Ultimately much of the understanding of physics comes from doing physics. It's a myth that you can understand a topic like this through pure thought and logic. It's only logical once you have done the calculations for yourself.

In any case, the change in KE is independent of reference frame. That's why you can't just magic away a change in energy by looking only at the car and ignoring the Earth. The Earth is a very large object to ignore!

If you put a car in free space it can't accelerate at all. So, if you really want to ignore the Earth then the car has zero acceleration, regardless of the engine's power.

 

[phystro]

Dont understand you,but never mind..
This topic is too complicated for my logic,problem is that I still can't understand why object is harder to accelerate if travel with higher speed.
In my brain any constant speed and zero speed is same thing,from car perspective/frame

You wouldn't need the Earth to understand that. You just need to consider two different frames of reference in relative motion. Assuming that you could magically accelerate the car in empty space without friction (and without propellant i.e utilising Newton's 3rd law): Initially you have a frame of reference where the car is at rest, however once the car reaches a speed of 100km/s, you would need to change a frame of reference to say that the car is at rest. These two frames of reference would have a relative speed of 100km/s, and hence what you considered to be a single car reference frame, are in reality two different frames in relative motion. Does this makes sense?

 

[Lnewqban]

Lets assume engine produce constant power all the time,to make things easier

 

[Jurgen M]

View attachment 291827

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..

 

[jack action]

Let imagine that car with constant 500HP accelerate [...]

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?

Pretty easy to analyze:

Power $P$ is 500 hp, or 372 850 W. Speeds are 100 km/h and 200 km/h, or 27.78 m/s and 55.56 m/s.

Energy spent: $E_s = Pt = 372\ 850\ W \times 4\ s = 1\ 491\ 400 J$

Therefore the amount of energy given by the 200 mL of petrol must also be 1 491 400 J.

The amount of kinetic energy of the car must also be 1 491 400 J. Or $\frac{1}{2}m(27.78^2 - 0^2) = 1\ 491\ 400 J$. This effectively tells us what the mass of the car is (i.e. 3865 kg).

What happens from 100 km/h to 200km/h?

The same amount of fuel spent? Therefore there is also 1 491 400 J of energy spent.
Same power? Therefore the time must also be the same (4 s) since the energy from the fuel is the same. Same mass? Therefore the kinetic energy must $\frac{1}{2}3865(v_f^2 - 27.78^2) = 1\ 491\ 400 J$ or the final velocity $v_f = 39.29\ m/s$ or 141 km/h.

So the same amount of energy spent at different speeds doesn't give the same $\Delta v$, but rather the same $\Delta(v^2)$.

Note also that if you increase the time then you must decrease the power to respect the amount of energy given by the fuel. And the final speed will still remain the same.

 

[russ_watters]

But why would object will be harder to accelerate when object travel at higher speeds if speed is relative term?
I don't understand this concept at all..

You've chosen a reference frame against which to measure speed. It has a physical reality/relevance to the car that means you can't arbitrarily change it later. Surely you must see that relevance/reality?

 

[Jurgen M]

You've chosen a reference frame against which to measure speed. It has a physical reality/relevance to the car that means you can't arbitrarily change it later. Surely you must see that relevance/reality?

You choose Earth ground as ref. frame when calculate kinetic energy. I am in the car engine,if I travel 0km/h ,100km/h or 2000km/s for engine doesn't make difference, he don't feel nothing .engine just feel acceleration when fuel consumption rise...

 

[Nugatory]

Lets assume engine produce constant power all the time,to make things easier

If the engine produces constant power, then the torque and hence the acceleration will not be constant.

In fact, we can't calculate the acceleration from the power at all - the power just tells us what the top speed will be when there are resistive forces such as friction and air resistance. To calculate the acceleration we need to know the relationship between engine RPM and torque (mostly based on the characteristics of the engine), between engine RPM and wheel RPM (because that gives us the torque at the wheels), and between torque at the wheels and forward force on the car (which is the only simple calculation - it just depends on the diameter of the tires).

 

[sophiecentaur]

e.g. throwing stuff out of the back of the car gives the same acceleration regardless of the speed relative to the Earth.

Going a step further gives you a rocket, which loses mass all the while so the equation becomes harder. A step even further could involve an Ion Drive thruster, which ejects a vanishingly small mass (ions from a particle accelerator) at extremely high velocity (near c) so the rate of increase of momentum can be regarded as almost constant.

So the main problem is that engine produce constant power not constant thrust?

Back to an 'ideal car' propulsion system and this is correct (ideal infinitely variable gearbox). Constant Power from the engine will give a constant rate of increase in Kinetic Energy with time. KE is proportional to the square of the velocity. So velocity will increase proportionally with the square root of the time.

 

[Jurgen M]

Pretty easy to analyze:

Power $P$ is 500 hp, or 372 850 W. Speeds are 100 km/h and 200 km/h, or 27.78 m/s and 55.56 m/s.

Energy spent: $E_s = Pt = 372\ 850\ W \times 4\ s = 1\ 491\ 400 J$

Therefore the amount of energy given by the 200 mL of petrol must also be 1 491 400 J.

The amount of kinetic energy of the car must also be 1 491 400 J. Or $\frac{1}{2}m(27.78^2 - 0^2) = 1\ 491\ 400 J$. This effectively tells us what the mass of the car is (i.e. 3865 kg).

What happens from 100 km/h to 200km/h?

The same amount of fuel spent? Therefore there is also 1 491 400 J of energy spent.
Same power? Therefore the time must also be the same (4 s) since the energy from the fuel is the same. Same mass? Therefore the kinetic energy must $\frac{1}{2}3865(v_f^2 - 27.78^2) = 1\ 491\ 400 J$ or the final velocity $v_f = 39.29\ m/s$ or 141 km/h.

So the same amount of energy spent at different speeds doesn't give the same $\Delta v$, but rather the same $\Delta(v^2)$.

Note also that if you increase the time then you must decrease the power to respect the amount of energy given by the fuel. And the final speed will still remain the same.

So conclusion is ,if you accelerate car at higher speeds that leads to higher fuel consumption even if resistance forces don't exist?
In real life there is resistance forces plus, so situation is even worser..

 

[jack action]

even if resistance forces don't exist?

Inertia is a resistance that cannot be avoided ($F = ma$).

 

[Jurgen M]

Inertia is a resistance that cannot be avoided ($F = ma$).

Yes I know that, mass inertia is only thing that count in my question.
inertia is not real force it is property.

 

[jack action]

Yes I know that, mass inertia is only thing that count in my question.
inertia is not real force it is property.

You still have to fight it when changing velocity (i.e. accelerating) and that requires energy.

 

[Jurgen M]

You still have to fight it when changing velocity (i.e. accelerating) and that requires energy.

I know that but what I didnt know is that accelerating at higher speeds costs me more at fuel consumption.

So for same amount of fuel if I start accelerating from 0km/h I will make delta V of 100km/h but if I start accelerate at 100km/h will make delta V only 41km/h...

hmm that is fascinating.

 

[russ_watters]

You choose Earth ground as ref. frame when calculate kinetic energy. I am in the car engine,if I travel 0km/h ,100km/h or 2000km/s for engine doesn't make difference, he don't feel nothing .engine just feel acceleration when fuel consumption rise...

No, the car and ground interact in a very real/non-arbitrary way to make the car move. You can't transform that away by changing reference frames.

 

[Lnewqban]

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.

I hope you have changed your take on the above statement.

https://en.wikipedia.org/wiki/Work_(physics)#Work–energy_principle

 

[Jurgen M]

Car wheel don't act at Earth c.g. so car has tendency to push Earth back and rotate,translation+ rotation ?
If car is very very big he will push Earth just like little stone when comes at tyer?

 

[A.T.]

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...

This is already unphysical. With no loses, any non-zero power input would produce an infinite propulsive force at velocity = 0. In reality the propulsive efficiency tends towards zero when the velocity relative to the propulsive medium (here ground) does.

 

[A.T.]

Car wheel don't act at Earth c.g. so car has tendency to push Earth back and rotate,translation+ rotation ?
If car is very very big he will push Earth just like little stone when comes at tyer?

View attachment 291839

The common CoM of planet and car will stay fixed. But the planet's CoM will oscillate as the car drives around it.

 

[Jurgen M]

This is already unphysical. With no loses, any non-zero power input would produce an infinite propulsive force at velocity = 0. In reality the propulsive efficiency tends towards zero when the velocity relative to the propulsive medium (here ground) does.

I started this topic to find out what is the main reason/contribution why car accelerate slower at higher speeds,because I always doubt that air can slow down acceleration so much.It must be something else but is hard to understand with pure logic/intuition because delta V is the same.

 

[PeroK]

Car wheel don't act at Earth c.g. so car has tendency to push Earth back and rotate,translation+ rotation ?
If car is very very big he will push Earth just like little stone when comes at tyer?

The calculations are more complicated if we consider the rotation of the Earth, so let's simpify things by imagining a car of mass $m$ pushing against a large object of mass $M$. The car has an engine which releases energy and creates relative motion between the mass $m$ and $M$. The engineering details are not relevant to show where your mistake is.

First, we look at the initial rest frame. The car moves in one direction at $v$ and the large object in the other direction at speed $V$. By conservation of momentum we have:$$mv = MV$$ Now, we calculate how much energy is needed:$$E = \frac 1 2 mv^2 + \frac 1 2 MV^2 = \frac 1 2 mv^2 + \frac 1 2 M(\frac{mv}{M})^2 = \frac 1 2 mv^2(1 + \frac m M)$$And if $M$ is much larger than $m$, we see that $\frac m M \approx 0$ and we have: $$E \approx \frac 1 2 mv^2$$ We also see that the amount of energy needed increases as $v^2$. E.g. if we want to car to travel at $2v$ relative to the large object, then we need $4$ times the energy.

Now, we try your trick of achieving a speed of $v$ and then moving to the rest frame of the car. What you say is that the car now has zero velocity again. But, what you forget is that in this frame of reference the large object now has speed $v$. The KE of the system in the car's rest frame is: $$E_1 = \frac 1 2 M v^2$$ And the initial momentum is $p_1 = -Mv$ (if we take the direction of the car to be positive). I just want to stress that you cannot simply ignore this, as you have been doing in all your posts.

In any case, we now want to increase the relative speed of the car and object to $2v$, which means approximately getting the car moving at speed $v$. Again, by conservation of momentum we have $$mv - M(v + V) = -Mv$$ and again we have $$mv = MV$$ Now the total KE of the system is:
$$E = \frac 1 2 mv^2 + \frac 1 2 M(v+V)^2 = \frac 1 2 mv^2 + \frac 1 2 Mv^2(1 + \frac m M)^2$$
$$E = \frac 1 2 mv^2 + \frac 1 2 Mv^2(1 + 2\frac m M + \frac{m^2}{M^2}) = \frac 1 2 mv^2 + \frac 1 2 Mv^2 + mv^2 + \frac 1 2 mv^2(\frac m M)$$ Again, the last term is approx zero and we have the change in energy: $$E - E_1 = \frac 1 2 mv^2 + mv^2 = \frac 3 2 mv^2$$ And that is the same change in energy as we found in the original rest frame, where $$E' - E'_1 = \frac 1 2 m(2v)^2 - \frac 1 2 mv^2 = \frac 3 2 mv^2$$
It doesn't matter whether you change reference frames of not, the change in kinetic energy is always the same. As it must be, as the energy came from the fuel consumption, which is independent of reference frame.

 

[PeroK]

PS there is also a very serious point that you can never understand physics if you don't want to get your hands dirty by doing calculations.

 

[Jurgen M]

PS there is also a very serious point that you can never understand physics if you don't want to get your hands dirty by doing calculations.

I agree but I can't do calculation if my logic is completely wrong..I didnt know how to start...
In car frame I forgot for Earth,I behave like Earth don't exist

 

[phystro]

I think it was confusing because of this detail (I now see that you must use u not du ---> P=Fu, as I mentioned in my first post):
A) The graph of Kinetic energy vs velocity wouldn't be a linear graph (it has already been posted). You would need 4 times more kinetic energy to achieve two times higher velocity, with the same acceleration.
B) The graph of Power vs velocity would be linear (P=Fu - for constant acceleration). This means that you would need exactly 2 times more power to achieve two times higher velocity.
C) This is explained because power is the time derivative of energy. i.e: $P= dE/dt$

 

[A.T.]

is hard to understand

The key relationship is quite simple:
power_engine * efficiency = power_propulsion = force * velocity

 

[meekerdb]

But why would object will be harder to accelerate when object travel at higher speeds if speed is relative term?
I don't understand this concept at all..

So you don't agree with comments from member @m4r35n357 in site below?
https://physics.stackexchange.com/q...ct-at-higher-speeds-if-speed-is-relative-term

It's because you have posed the problem in terms of constant power, where power is a relative value, because kinetic energy is a relative value. If the car had a rocket engine it could have constant thrust and acceleration which are absolute values.

 

[meekerdb]

The easy solution is to equate energies: 1/2 mv^2 = Pt so v = sqrt[2Pt/m].

 

[Jurgen M]

It's because you have posed the problem in terms of constant power, where power is a relative value, because kinetic energy is a relative value. If the car had a rocket engine it could have constant thrust and acceleration which are absolute values.

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?

 

[A.T.]

how explain this in reference frame of rocket without violate physics laws?

Do the calculation in both frames and compare. But don't forget to include the kinetic energy of the expelled exhaust gases.