# Does car use more fuel when accelerating (for same delta) at higher speeds if resistance forces are constant?

If air and tyre resistance are constant over speed(unrealistic), does car use same time,same total fuel,same fuel/ distance and same fuel/time to accelerate from 100 to 150km/h then from 200 to 250km/h?

Assume in both case engine works in same RPM range (continuos variable transmission), so engine efficiency/power is the same.

• Power is force $\times$ velocity. Sep 30, 2023 at 10:05
• @Farcher What "push car back" more from 200-250km/h then 100-150km/h if resistant forces are the same? Sep 30, 2023 at 10:19
• But the speeds are not so that rate of working, rate of consumption of fuel, changes. Put another way, work done = force $times$ distance, so compare the distances travelling during the acceleration phases. Sep 30, 2023 at 10:44
• But air and tyre resistance aren’t constant over speed. You can prove anything by assuming a counterfactual. Sep 30, 2023 at 10:54

As somebody has already pointed out $$$$P=\vec{F}\cdot\vec{v},$$$$ therefore, even assuming zero friction, if you fix the power of the engine, then the force (and the acceleration, by Newton's second law, as well) decreases as your velocity increases; the same would happen even assuming constant forces like the type of friction that you described. That being said, your approximation is highly unrealistic. Air resistance depends on the velocity; in the turbulent regime, where typically the aerodynamic drag reads $$F_D=bv^2$$, where $$b$$ does not depend on $$v$$, with the assumption of fixed power and neglecting tyre resistance, this even tells you that the maximum speed is determined by the power of your engine: $$$$F_D=F_E\Rightarrow P=F_{max}v_{max}=bv_{max}^3\Rightarrow v_{max}=\sqrt[3]{\frac{P}{b}},$$$$ where $$F_E$$ represents the force applied by the engine.

Regarding the fuel consumption, assuming that the power provided by your engine is constant, you need less time to accelerate from 100 to 150 km/h than from 200 to 250 km/h. So the total amount of energy ($$E=P\Delta t$$) used in the former case is less than in the latter. Assuming ideal efficiency in both the cases, the energy is proportional to the amount of fuel, therefore less total amount of fuel is needed in the former case. Fuel per time, under the above assumptions, is instead constant, because it is proportional to the power exerted by the engine, which we assumed to be constant.

• Are total amount of fuel, fuel per distance, fuel per time same in both case? Is time also same for both case Sep 30, 2023 at 15:53
• I thought that editing my answer would be a better idea than replying with a long comment. Sep 30, 2023 at 19:13
• you forgot write fuel per distance Oct 2, 2023 at 14:44
• You can write the fuel per distance as dW/ds=dW/dt*dt/ds=dW/dt*1/v, and, since dW/dt is constant, this means that fuel per distance decreases as velocity increases; this is reasonable if you have zero friction: if your velocity increases, and your power stays constant, then in the same amount of time, with the same amount of fuel, you travel a larger distance. Assuming constant friction F_D, of course, there is a maximum velocity given by v_max=P/F_D. Oct 5, 2023 at 23:45

1. fuel / time (L/h) is equal in both case

because engine works in both case at same power/full throttle

2.From 200-250km/h acceleration is lower, time is longer

because engine has constant power(P) but thrust(F) drops as speed rise, this can be seen in two ways:

a) P = F x v (to keep P constant, F must drops when v rise)

b)if variomat keep same RPM as speed rise, torque(thrust(F) x radius) at wheel drops, because gear ratio drops.

3. fuel / distance (L/100km/h) is lower for case 200-250km/h

because car travel faster so it make greater distance in this run.

4. Total amount of fuel is higher for case 200-250km/h

because time that car travel at full power is longer then in first case. we can also see this from kinetic energy E= (m x v*2) / 2