Kinematics Acceleration question

[Amad27]

Homework Statement

Suppose a can, after an initial kick, moves up along a smooth hill of ice. Make a statement concerning its acceleration.
A) It will travel at constant velocity with zero acceleration.
B) It will have a constant acceleration up the hill, but a different constant acceleration when it comes back down the hill.
C) It will have the same acceleration, both up the hill and down the hill.
D) It will have a varying acceleration along the hill.

Homework Equations

N/A

The Attempt at a Solution

I thought the answer is D) It will have a varying acceleration along the hill

So, it has an initial kick which means $v_0 \ne 0$
$v_0 > 0$
$x_0 = 0$

Now, an object moving has varying acceleration if $x(t) = at^3 + bt^2 + ct + d$ third degree polynomial.

Can the motion be modeled by $x(t) = t^3$? ?

They say the answer is (C) but I believe it is D

 

[FaroukYasser]

Since the question said that it is moving up a smooth hill, it means that the friction force is negligible therefore the only force acting on the can affecting its acceleration is its weight.
It the Slop the can is moving upwards in is for example at an angle of 30 degrees then we can say that:
mgsin30 = ma
a = mgsin30/m
a = gsin30

Going down, it is the same thing, only weight acts on it so the acceleration is gsin30 again.
Generally in such examples (only weight acting on an object on a slope) you can quote the acceleration as gsin(theta) where theta is the angle which the slope makes with the horizontal line.

 

[Amad27]

ince the question said that it is moving up a smooth hill, it means that the friction force is negligible therefore the only force acting on the can affecting its acceleration is its weight.
It the Slop the can is moving upwards in is for example at an angle of 30 degrees then we can say that:
mgsin30 = ma
a = mgsin30/m
a = gsin30

Going down, it is the same thing, only weight acts on it so the acceleration is gsin30 again.
Generally in such examples (only weight acting on an object on a slope) you can quote the acceleration as gsin(theta) where theta is the angle which the slope makes with the horizontal line.

Cant the position be modeled by a cubic function $x(t) = ax^3 + bx^2$??

 

[FaroukYasser]

I don't understand why you are trying to model the position of the can when you were not asked. For all you know, its position can be modeled by any function. In this case the function would be linear (degree 1 polynomial) so the change in displacement is constant hence the acceleration is constant.

This example is exactly the same as a model for free fall where the only force acting on a falling object is its weight. In this example (and your example) we can use linear regression to model the position of an object.

 

[Amad27]

I don't understand why you are trying to model the position of the can when you were not asked. For all you know, its position can be modeled by any function. In this case the function would be linear (degree 1 polynomial) so the change in displacement is constant hence the acceleration is constant.

This example is exactly the same as a model for free fall where the only force acting on a falling object is its weight. In this example (and your example) we can use linear regression to model the position of an object.

Right, so it can be modeled as a a function, suppose -x^4.

Because you can model it as that function.

$x(t) = t^4$
$v(t) = 4t^3$
$a(t) = 12t^2$

Which means it could have definitely have a varying acceleration.

 

[FaroukYasser]

You can model it as a function.. A linear function. You are modelling it with a non linear function.

A linear function is a function that has a polynomial of first degree.

 

[Amad27]

You can model it as a function.. A linear function. You are modelling it with a non linear function.

A linear function is a function that has a polynomial of first degree.

It can't be a line because

Suppose the peak of the position. The exact top until the can starts to go down is considered t = 0 then

x(-a) = x(a) has to be s true equation. Which can only be possible if x(t) is nonlinear

 

[FaroukYasser]

The function that you use to model tells you the position of the ball at any time. So the start is t = 0 and at the top the time = 5 s for example and at the bottom the time is 10 seconds. Substituting the time in the function gives you the place of the can.

I don't understand why you are very committed to the "modelling using a function" idea. Very very basic mechanics can answer this question in no time since there is not a load of variables involved.

Did you think about it using mechanics?

 

[Amad27]

The function that you use to model tells you the position of the ball at any time. So the start is t = 0 and at the top the time = 5 s for example and at the bottom the time is 10 seconds. Substituting the time in the function gives you the place of the can.

I don't understand why you are very committed to the "modelling using a function" idea. Very very basic mechanics can answer this question in no time since there is not a load of variables involved.

Did you think about it using mechanics?

Because using mathematics is more "proof" of something.

The position CAN be modeled by x(t)=x^4 Considering t=0 is the top place of the ball (before it comes down) and the initial position x0 is the peak (before ball comes down)

 

[FaroukYasser]

In this example, you can't model the can except in a Linear model since it is not affected by any varying forces. The force on the can is constant, the force on it is always weight which is constant. You can use non linear models if on the other hand the forces were varying. For example a ball falling in air with air resistance since the resisting forces would be increase as you move.

Finally if you want a math proof... think about your model for a moment
x(t) = t^4
so at time 0s, the position of the can is 0
at time 1s , the position is 1
That is an acceleration of (1-0)/1 = 1 ms^-2
Next
at time 2s, the position is 8
That is an acceleration of (8 - 1)/2 = 7/2 ms^-2

using YOUR model, the acceleration of the object has increased.
But tell me how can acceleration increase without increasing the force on the object?
We already agreed that the only CONSTANT force acting on the can is Weight. how did weight miraculously increase to cause this increase in acceleration??
Hence I just disproved your model.

Doing the same with any linear model like for example:
x(t) = 2t
at time 0, position is 0
at time 1, position is 2
acceleration = (2-0)/1 = 2 ms^-2
Next
at time 2, position is 4
Acceleration = (4 - 2)/1 = 2 ms^-2
and so on the acceleration would always remain of constant value of 2 ms^-2 since the slope of the function is constant.
and this is logical because since the Weight force is constant then the acceleration is also constant.
Force is directly proportional to acceleration. Anything that happens to one happens to the other

And by the way proving that the acceleration is constant using mechanics is also a proof. You don't have to involve "mathematics" in.

 

[Amad27]

I'll

In this example, you can't model the can except in a Linear model since it is not affected by any varying forces. The force on the can is constant, the force on it is always weight which is constant. You can use non linear models if on the other hand the forces were varying. For example a ball falling in air with air resistance since the resisting forces would be increase as you move.

Finally if you want a math proof... think about your model for a moment
x(t) = t^4
so at time 0s, the position of the can is 0
at time 1s , the position is 1
That is an acceleration of (1-0)/1 = 1 ms^-2
Next
at time 2s, the position is 8
That is an acceleration of (8 - 1)/2 = 7/2 ms^-2

using YOUR model, the acceleration of the object has increased.
But tell me how can acceleration increase without increasing the force on the object?
We already agreed that the only CONSTANT force acting on the can is Weight. how did weight miraculously increase to cause this increase in acceleration??
Hence I just disproved your model.

Doing the same with any linear model like for example:
x(t) = 2t
at time 0, position is 0
at time 1, position is 2
acceleration = (2-0)/1 = 2 ms^-2
Next
at time 2, position is 4
Acceleration = (4 - 2)/1 = 2 ms^-2
and so on the acceleration would always remain of constant value of 2 ms^-2 since the slope of the function is constant.
and this is logical because since the Weight force is constant then the acceleration is also constant.
Force is directly proportional to acceleration. Anything that happens to one happens to the other

And by the way proving that the acceleration is constant using mechanics is also a proof. You don't have to involve "mathematics" in.

Exactly my point.

You DONT know all the forces acting on the can. Perhaps a wind?

Also, my second point, MATHEMATICALLY, x(t)=x^4 COULD, doesn't have to be the position GRAPH.

 

[FaroukYasser]

"Suppose a can, after an initial kick, moves up along a smooth hill of ice"
You have to infer that by saying can, he means something small that you neglect the air resistance, + air resistance is affected by the speed of the object .and a simple kick (or even a hard one) wouldn't make air resistance

Also, There are NO other forces acting on the can except the weight. Name me any other force (other than air resistance since I just told you it is negligible and other than friction since ice has almost no friction)

If you could name me one other force AFFECTING the acceleration of the can then truly you can model it as a non linear function.
But there simply isn't any other force. The only SIGNIFICANT force acting is weight.

Finally, what do you mean it doesn't have to be a position graph? your modeled function :
x(t) = a non linear function of t
is a model of position. you substitute in values for time and it tells you the position of the can at that exact time. so IT IS a position graph.

Also, one more thing, You typically can model any type of function for a real life situation and expect it to work: if I model the function:
x(t) = sin
then I must have proof that the function is true BEFORE I apply calculus to it. How do you know that your function is true or not?
You just assumed a random function (that is incorrect) and applied you knowledge of calculus to the wrong function resulting in wrong results.

 

[Amad27]

"Suppose a can, after an initial kick, moves up along a smooth hill of ice"
You have to infer that by saying can, he means something small that you neglect the air resistance, + air resistance is affected by the speed of the object .and a simple kick (or even a hard one) wouldn't make air resistance

Also, There are NO other forces acting on the can except the weight. Name me any other force (other than air resistance since I just told you it is negligible and other than friction since ice has almost no friction)

If you could name me one other force AFFECTING the acceleration of the can then truly you can model it as a non linear function.
But there simply isn't any other force. The only SIGNIFICANT force acting is weight.

Finally, what do you mean it doesn't have to be a position graph? your modeled function :
x(t) = a non linear function of t
is a model of position. you substitute in values for time and it tells you the position of the can at that exact time. so IT IS a position graph.

Also, one more thing, You typically can model any type of function for a real life situation and expect it to work: if I model the function:
x(t) = sin
then I must have proof that the function is true BEFORE I apply calculus to it. How do you know that your function is true or not?
You just assumed a random function (that is incorrect) and applied you knowledge of calculus to the wrong function resulting in wrong results.

You can't know what the position graph will be, you can SPECULATE the graph and equation.

So MATHEMATICALLY, there are infinite position graphs.

Wind is a natural force. Which could be acting on the can

 

[haruspex]

It is reasonably obvious that the question intends you to ignore friction, air resistance, variations in gravity... More seriously, it ought to state that the hill surface is a straight line.
But suppose you argue that it might not be straight line. Now none of the answers are correct. You can't say that the acceleration will vary, and you can't say that it won't. If you accept the set of answers as additional clarification, C is the only possible answer.