Car acceleration if resistance forces don't exist

In summary, the car will accelerate from 100-200km/h in 4 seconds and burn 200mL of petrol, just like it did from 0-100km/h, because the change in velocity is the same in both cases. However, the acceleration will be smaller due to the decrease in thrust caused by the engine producing constant power. This decrease in acceleration will result in the car taking longer than 4 seconds to reach 200km/h and burning more than 200mL of petrol. Therefore, the acceleration is not constant, as the power of the engine decreases with increasing speed. Additionally, the reference frame of the road is irrelevant in this case, as energy is frame-dependent. When analyzing the car-Earth system
  • #1
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?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
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.
 
  • Skeptical
  • Like
Likes bhobba, Delta2 and hutchphd
  • #3
Lnewqban said:
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
 
  • #4
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.
 
  • Skeptical
Likes nasu and PeroK
  • #5
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.
 
  • Like
  • Wow
Likes bhobba, Delta2 and nasu
  • #6
phinds said:
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.
 
  • #7
phystro said:
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
 
  • #8
Jurgen M said:
Point of question is important.
Accurate statement of the problem you are actually trying to solve is ALSO important.
 
  • Like
Likes sophiecentaur
  • #9
Jurgen M said:
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.
 
  • Like
Likes docnet, sophiecentaur, russ_watters and 2 others
  • #10
phinds said:
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.
 
  • #11
PeroK said:
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?
 
  • #12
Jurgen M said:
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.
 
  • Like
Likes nasu, docnet and Lnewqban
  • #13
@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.
 
  • Like
Likes cjl, russ_watters and PeroK
  • #14
jbriggs444 said:
@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
 
  • #15
Jurgen M said:
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.
 
Last edited:
  • Like
Likes PeroK
  • #17
jbriggs444 said:
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?
 
  • #18
Jurgen M said:
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.

Jurgen M said:
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.
 
  • #19
PeroK said:
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?
 
  • #20
Jurgen M said:
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.
 
  • #21
PeroK said:
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
 
  • #22
Jurgen M said:
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.
 
  • Like
Likes nasu and sysprog
  • #23
Jurgen M said:
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?
 
  • #24
Jurgen M said:
Lets assume engine produce constant power all the time,to make things easier
kinetic_energy_in_moving_car.png
 
  • #25
Lnewqban said:
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..
 
  • #26
Jurgen M said:
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.
 
  • Like
Likes Lnewqban
  • #27
Jurgen M said:
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?
 
  • #28
russ_watters said:
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...
 
  • Skeptical
Likes PeroK
  • #29
Jurgen M said:
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).
 
  • Like
  • Love
Likes bhobba and russ_watters
  • #30
PeroK said:
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.
Jurgen M said:
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.
 
  • Like
Likes bhobba
  • #31
jack action said:
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..
 
  • #32
Jurgen M said:
even if resistance forces don't exist?
Inertia is a resistance that cannot be avoided (##F = ma##).
 
  • #33
jack action said:
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.
 
  • #34
Jurgen M said:
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.
 
  • #35
jack action said:
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.
 

Similar threads

Replies
15
Views
2K
  • Classical Physics
Replies
4
Views
1K
  • Classical Physics
2
Replies
53
Views
3K
Replies
12
Views
670
  • Introductory Physics Homework Help
Replies
5
Views
973
  • Mechanical Engineering
Replies
18
Views
4K
  • Mechanics
Replies
10
Views
1K
  • Engineering and Comp Sci Homework Help
2
Replies
40
Views
3K
Replies
23
Views
6K
  • Aerospace Engineering
Replies
8
Views
2K
Back
Top