Friday, December 16, 2011

Car On A Hill

OK, I found a physics education paper that studied Intro Physics students understanding (or misunderstanding) of the concept of velocity, acceleration, and force in one dimension. One of the questions they used to test a student's understanding is actually quite interesting in the sense that it DOES appear to test how well a student actually understands the difference between velocity, acceleration, and force. So I thought, before I give the link to the paper, that I will ask the question here. If you are a student, or even just someone trying to learn physics, see if you can answer this:

A car is on a hill and the direction of its acceleration is uphill. Which statement best describes the motion of the car at that time?

A. it is moving uphill
B. it is moving downhill
C. it is not moving
D. both A and B are possible
E. both A and C are possible
F. A, B, and C are possible

Try it.

This is one example where one needs to understand something beyond just a superficial level. Many people will tend to pick the obvious answer because, well, it's obvious. But to understand why the correct answer is the correct answer will require an intimate knowledge of what velocity, acceleration, and force mean, and their relationships to each other beyond just a hand-waving understanding.

I'll give this a few days, and I'll edit this post to link to the paper in question. If you happen to have read the paper already, or better yet, one of the authors, please hold off your comment and let others try it first. Please post your answer on here, but I will hold off on releasing all comments with answers till AFTER a few days, so that no one will be influenced by any of the submitted responses. Comments that do not contain the answers will be released as usual.

EDIT: I'm getting a few responses already. Again, just a reminder, I'll keep the comments that contain answers moderated for now. I'll release those in a few days when I post a link to the paper. So if you don't see your comment appearing after you submit it, you'll know why.

EDIT (12/20/2011): I've posted the answer and the source paper that this question came from. Thanks to all those who participated and posted their answers.

Zz.

15 comments:

Blaise Pascal said...

The answer is F: it could be moving uphill, downhill, or not at all.

Curtis Faith said...

Could be any of the above so answer F is correct.

A, B, and C are questions of velocity. Acceleration is independent of velocity even though it is the rate of change of that velocity.

Consider a car moving backwards down a hill by gravity. If the person in the car accelerates forward using a constant force, the car will slow down and then start moving forward. During that time the car will transition between all three states defined by A, B, and C.

silent.alias said...

Acceleration is the derivative of the velocity with respect to time. If the car is currently stationary, it may be inflecting between having been rolling backward and is about to creep forward - as often happens with cars with stick shifts when positioned up a hill at a light. The acceleration is forward, but the car is stationary at that moment in time.

Forward movement is obvious.

Backward movement is also possible - if the car were rolling backward when the driver stepped on the gas. The change in velocity forward would be positive, since the change in velocity backward would be negative. The acceleration in this case is positive.

SFG said...

The answer is F.

Joey Dumont said...

The answer is F. The direction of the acceleration is uphill, but nothing is said as to its current velocity. If the car's instantaneous velocity is downhill (call this the negative direction) and it is accelerating uphill, it will eventually stop moving in the negative direction and its velocity will be positive. Hence, all the situations are possible.

Gandalf said...

The correct answer is F.

The car is on a hill and thus is assumed to be influenced by the force of gravity; in the direction of movement this force would be: sin(instantaneous angle)*m*g (I don't assume a constant sloped hill, but it would work too).

Anyway we're told that the car is actually accelerating uphill. This means that the engine of the car is providing a force opposite to- and of greater magnitude than the force of gravity. BUT we remember that acceleration is the second derivative of position, and first derivative of velocity. This means that the car could be having a velocity vector pointing downhill or uphill at the instant of the situation.

Therefore, depending in which direction the cars velocity has, the car could be speeding up (if already moving uphill) or speeding down (if moving downhill but having the acceleration uphill), OR it could be at the turning point (just went downhill but is now at the turning point just before moving uphill).

Note that when we say "it is moving ...", we're talking about the velocity. Knowing something about the acceleration tells of nothing of the existing velocity at this point, only how it's changing. This is actually a great point because it highlights exactly that Newtons 2. law is a 2nd order differential equation - we need suitable boundary conditions to completely know the motion of the object (for example the starting position and velocity).



--------In general-----------
And that's how most of our physical laws work: they describe how things CHANGE not how things ARE. We have to put this knowledge of how things ARE at some instant into the equations ourself as boundary conditions to pick out the correct solution to our problem out of the infinitely many solutions that the (1st or 2nd) order (ordinary or partial) differential equations give.
EXAMPLE: When we integrate acceleration with respect to time, we get a*t + c - this constant c can take any value and it'll still be a solution to dv/dt = a. But for the knowledge of a particular c (which we can call v0 since it's the starting velocity at t=0), we get THE ONE specific solution for v. An object starting with a different velocity at c but experiencing the same accceleration will too satisfy dv/dt = a, but will be another solution of the "function family" a*t + c. When we integrate acceleration with respect to time twice we get the wellknown x(t) = x0 + v0*t + 0.5*a*t^2, where we need the two mentioned boundary conditions to pick out a specific solution in the pool of infinite solutions where x0 and v0 take all possible values.

Tommy said...

Is gravity considered in this question?

ZapperZ said...

It is, or else having the car on a hill makes no difference! Why would it stick to the hill if there's no gravity? But read again the question, and consider the NET effect being mentioned.

Zz.

Deddryk said...

F is the answer. The question only deals with acceleration not velocity.

enderw88 said...

I think the obvious answer is 'F', that a, b, or c are equally good. But this ignores that the question specified a 'Car'. Cars do not have effective means of accelerating through a standstill. So while moving down hill actuating the brakes will give 'b', you cannot continue that activity such that 'c' would be momentarily true, so the acceleration up hill would cease, violating the terms of the question...

SO the correct answer is 'D'

George said...

The answer is F. For instance look at a car rolling down the hill backwards. Now imagine hitting the gas. After hitting the gas the car will slow to a stop and then change direction. Thus the car has acceleration up the hill (after hitting the gas) and yet, at some point in time, it is moving down the hill, up the hill and is not moving. Thus, the answer has to be F.

Alabanda said...

D. both A and B are possible

It may be moving downhill while decelerating, or moving uphill while accelerating.

(Having just learnt rolling motion, I wouldn't be surprised if no movement were also possible. Put the car on a icy hill! Ha.)

Anders Eg said...

F
Anders Eg

Anders Eg said...

F
Anders Eg

Deddryk said...

@enderw88

If it is not possible for the car to be accelerating upwards from a stopped position than it is not possible for the car to go up once it has stopped. There is a point at which the car has 0 velocity but is accelerating upwards.