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A car is moving up a hill at a constant velocity. Is the car accelerating?

I know it's simple, but I argued for an hour about this.

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A car is moving up a hill at a constant velocity. Is the car accelerating?

I know it's simple, but I argued for an hour about this.

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Doc Al

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Note that many arguments are really disagreements about the meanings of words. In physics, terms like

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Hootenanny

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No constant velocity = no acceleration.byee614 said:constant velocity.

-Hoot

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My friend argues that the car must be accelerating to counteract the force of gravity. He says that the force applied on the car when you press the gas pedal implies acceleration because F=ma.

I countered this arguement by saying you have to encorporate every force acting on the car when you use F=ma. There are 3 forces acting on the car. Gravity, the ground pushing on the car perpendicular to the ground, and the force when you press the gas pedal. The NET FORCE is zero. Thus F=ma means that 0=m(0) and acceleration is zero.

But he won't listen.

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Doc Al

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You are correct; your friend is not.

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Hootenanny

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In my opinion you are correct on your argument, it is thebyee614 said:I countered this arguement by saying you have to encorporate every force acting on the car when you use F=ma. There are 3 forces acting on the car. Gravity, the ground pushing on the car perpendicular to the ground, and the force when you press the gas pedal. The NET FORCE is zero. Thus F=ma means that 0=m(0) and acceleration is zero.

-Hoot

[Edit] Doc Al is still faster than me

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vanesch

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And to be pedantic and confusing:Hootenanny said:No constant velocity = no acceleration.

In *Newtonian* physics, this is correct (if we consider the reference frame with which the velocity is defined, is considered sufficiently inertial).

In *General relativity*, the car is "accelerating away" from its geodesic with about the earth's gravity acceleration, g.... as does a car that is standing still, btw.

I'm saying this because it might be THIS which was at the origin of the discussion ; that using the equivalence principle, that a car at constant velocity (whether it is going uphill or downhill or on a flat road) is undergoing the force of gravity, which is equivalent to AS IF IT WERE ACCELERATING (and "pulling 1 g")

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russ_watters

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To help him understand why he is wrong, ask him if a book sitting on a table is accelerating. The table is exerting an upward force on it...byee614 said:He says that the force applied on the car when you press the gas pedal implies acceleration because F=ma.

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Doc Al

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Ok, I showed my friend this thread and he says that this statement about "general relativity" is what he's trying to argue. Which I think is BS because there was no mention of relativity or frame of reference when we first started arguing about it. It was a problem given on a Physics 101 college midterm.vanesch said:And to be pedantic and confusing:

In *Newtonian* physics, this is correct (if we consider the reference frame with which the velocity is defined, is considered sufficiently inertial).

In *General relativity*, the car is "accelerating away" from its geodesic with about the earth's gravity acceleration, g.... as does a car that is standing still, btw.

I'm saying this because it might be THIS which was at the origin of the discussion ; that using the equivalence principle, that a car at constant velocity (whether it is going uphill or downhill or on a flat road) is undergoing the force of gravity, which is equivalent to AS IF IT WERE ACCELERATING (and "pulling 1 g")

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YOu are right. Your friend is giving you a bunch of BS.byee614 said:Ok, I showed my friend this thread and he says that this statement about "general relativity" is what he's trying to argue. Which I think is BS because there was no mention of relativity or frame of reference when we first started arguing about it. It was a problem given on a Physics 101 college midterm.

Constant velocity does not mean there aren't any forces acting on an object. It just means that the vectorial SUM of all these forces produces a net force of ZERO, thus, no acceleration. The car must produce some force to counteract the force of gravity. The presence of such forces does not automatically imply an acceleration.

If he wants to argue about the equivalence of "gravity" and "acceleration", tell him he should at the very least WAIT till he actually understands basic newtonian kinematics before jumping rear-end first into general relativity.

Zz.

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cepheid

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Edit: nevermind. He specified that the velocity was constant from the very beginning of the thread. So I guess the hill is a ramp, by assumption.

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If the hill is a ramp, with Constant slope, then for the time that the car is moving on a ramp with a constant slope then the acceleration is zero..

But if the hill doesn't have a constant slope, or if you are thinking of before the car starts on the hill, till the time its on the ramp, the car had to accelerate when changing its slope....

and dont worry about curvature of earth or anything. cause as long as the slope is constant, then the cars path is constantly straight in the universe....

basically

constant slope = constant velocity

variable slope = variable velocity ->accerelation

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berkeman

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Hootenanny

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:rofl: :rofl: :rofl: :rofl:berkeman said:Centri-something acceleration....

That made me giggle. Just to defend myself the OP did state constant velocity, and in circular motion velocity isn't constant.

-Hoot:rofl:

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berkeman

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Good point. So that means the OP's friend wins the bet after all, since it's pretty hard to make a car that can sustain a constant velocity of 24,000 mph westward....Hootenanny said::rofl: :rofl: :rofl: :rofl:

That made me giggle. Just to defend myself the OP did state constant velocity, and in circular motion velocity isn't constant.

-Hoot:rofl:

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russ_watters

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Sorry, you're still wrong. If the problem defines a constant velocity, then the velocity is constant. Period.berkeman said:Good point. So that means the OP's friend wins the bet after all, since it's pretty hard to make a car that can sustain a constant velocity of 24,000 mph westward....

Besides, in this case, all a constant velocity assumption requries is the assumption of an earth-based reference frame - and that is not only reasonable, it is necessary for the problem to be clear and meaningful.

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vanesch

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That was the point I tried to make, but admittedly when talking about a car and a hill and so on, one can take it that it is seen in the framework of Newtonian gravity and not in general relativity...PatPwnt said:

Again, in the Newtonian framework, gravity is a FORCE, and space is Euclidean. As such, no matter what is going on, if the problem states that VELOCITY IS CONSTANT, then, acceleration being the derivative of velocity, and the derivative of a constant being zero, acceleration is zero.

And if you're now talking about "in which frame", well, I think it is fair to say IN THE SAME FRAME AS THE ONE WHERE THE VELOCITY WAS DEFINED.

So there's no discussion: in Newtonian physics, if velocity (in frame A - whatever it is, inertial or not!) is constant, then acceleration (in the same frame A, inertial or not) is zero.

Even if frame A is a non-inertial frame, btw. There will be "inertial forces" and all that working upon the object, but that's not the point. The point is:

acceleration = derivative of velocity

Always.

The link with forces is something else.

Now, in general relativity, things are different. Gravity is not a force, but a curvature of spacetime, and as such, the notion of "derivative" becomes a bit more involved. And then it is fair to say that the best "geometrical" way to define acceleration is by its "deviation from a geodesic", which would, for a car in said situation, correspond to about the acceleration of gravity.

However, such a car would not have "constant velocity as a geometrical object" either! At most one could talk about "constant derivatives of spatial coordinates wrt to the time coordinate in a specific reference frame". And in as much as THIS is accepted as a definition of "velocity", then we're back to the original statement: if we now take it that "acceleration" is the second derivatives of the spatial coordinates wrt the time coordinate in a specific reference frame", then, for the same reason as before: if the FIRST derivative is constant, then the SECOND derivative is zero.

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