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Based on the understanding that Newtons third law states for every action there is an equal and opposite reaction, when a car moves forward under driving force, air resistance and friction act equally in the opposite direction, leading to the car moving at a constant speed with no resultant force.

If this is the case, how is it possible for a car to accelerate?

Presumably the only way for a car to accelerate would be for the force acting in the forwards direction to be greater than the forces acting opposite to it. How would this be possible if the forces acting in the opposite direction must also be equal?

Is it true that air resistance etc takes time to become equal to the force driving the car forward, so by constantly increasing the force in the forwards direction the opposite force is unable to equal it simultaneously?

I am not sure this explanation is correct, so I would appreciate some help on the matter!

*To clarify, I can understand movement if you separate it from acceleration, in that the force from the ground is equal and opposite to the tyres, causing the car to move forward, the problem arose with the presumption that air resistance must equal the force the earth is exerting on the car, and not the force the car is exerting on the air. The question flagged as a duplicate had a similar problem but I believe with a significantly different example and slightly different line of questioning.

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marked as duplicate by Kyle Kanos, JMac, PM 2Ring, GiorgioP, ZeroTheHero Apr 11 at 12:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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That's not quite how Newton's Third Law works.

Newton's third law describes an action-reaction pair, which for our purposes here is a pair of interacting objects. Let's call these objects A and B. By "interacting" we mean that A is exerting a force on B, and B is exerting a force on A. It's also crucial to keep in mind that objects respond only to forces exerted on them, not forces exerted by them. What Newton's Third Law says is that the force exerted on A by B is equal and opposite to the force exerted on B by A.

Let's start with the simplest case, with no air, just the car and the ground. In this case, the action-reaction pair is the car('s wheels) and the ground. A car's engine exerts a force to rotate the car's wheels. The wheels, being in contact with the ground and subject to static friction, exert a force on the ground. In accordance with Newton's Third Law, the ground exerts an equal and opposite force on the car. This reaction force on the car is what causes the car to accelerate.

Now, if we add in the air, there are three possible action-reaction pairs: the car and the ground (which we already covered), the car and the air, and the ground and the air (which isn't strictly relevant to what the car does*). As we said before, the car's wheels exert a force on the ground, and the ground exerts an equal and opposite force on the car. The car accelerates due to the force exerted on it by the ground. As this is happening, let's consider the interaction of the car and the air. The air is in the way of the car, so as the car accelerates, it pushes on the air, exerting a force on it. This force on the air causes the air in front of the car to accelerate. By Newton's Third Law, the air exerts an equal and opposite force on the car, which is what air resistance is. What's crucial to keep in mind here is that this is a different action-reaction pair than the car-ground pair, so the force exerted on the car by the air does not have to be the same as the force exerted on the car by the ground. This is why the car can accelerate - the force exerted on the car by the air is generally much less than the force exerted on the car by the ground (and conversely, the force exerted by the car on the air is much less than the force exerted by the car on the ground), except at very high speeds, where there is quite a lot of air to push out of the way per second.

*If you're curious, the air-ground action-reaction pair works like this: the air has mass, and is pulled down by gravity, so its weight exerts a force on the ground. The ground, in turn, supports the weight of the air by exerting an upward force on the air.

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  • $\begingroup$ What about in the example of a skydiver? Are gravity acting on the skydiver and the skydiver acting on the air different pairs and that's why the skydiver is able to accelerate at first? $\endgroup$ – Jacob Apr 8 at 19:14
  • $\begingroup$ @Jacob Precisely. If we ignore the gravitational force between you and the air (which is so tiny that it can safely be ignored), then there are two relevant action-reaction pairs: you and the air, and you and the Earth. The Earth pulls you downward via gravity, and you pull the Earth upward with an equal and opposite force, also via gravity. You push the air out of the way, and the air pushes back on you with an equal and opposite reaction force. Once again, the forces belong to two different action-reaction pairs, so they can be different (and, until you hit terminal velocity, they will be). $\endgroup$ – probably_someone Apr 8 at 19:35
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This is a basic misconception of Newton's third law. The force that an object exerts receives an equal but opposite reaction force. Yes, but those forces act on different bodies!

If you puosh a wall, you exert a force on the wall, and the wall exerts a force on you. So you are not suffering your own force, only the wall's force on you. That's why you can do push ups.

So the forces do not cancel each other because they act on separate bodies.

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Newton's three laws build on each other and you are quoting the last "for every action there is an equal and opposite reaction". You are also alluding to the first, which is that objects in motion tend to stay at that same motion.

The answer to your question requires the second law. When the car's tires try to spin at a faster rate the force of friction tries to propel the car forward at an increased speed. In order to push back the car must generate a backward force which is does by accelerating. So, $F = m a$, and the car has to accelerate in order to generate the "push back" force.

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