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Class Notes Physics I, 9/09/98

Uniformity of acceleration on a constant slope; Equations of Uniformly Accelerated Motion



We attempt to determine by an experiment whether the acceleration of a friction cart on a constant incline depends on its velocity or its position on an incline.  We then derive the four equations of uniformly accelerated motion.   Finally we consider the difference between the effect on final velocity of a given acceleration through a distance, with different initial velocities, with its effect over a time interval with different initial velocities.

Introduction, Goals and Questions

Having defined acceleration, we attempt to establish at least one set of conditions under which acceleration should be uniform, namely a car coasting down a constant incline.  We take data for later analysis.

We then formulate the consequences of uniform acceleration, using

We finally see that from uniform acceleration and time interval, we can determine change in velocity while from uniform acceleration and distance we determine the change in the squared velocity.

The following questions arise:

Experiment:  Uniformity of acceleration on a constant slope

We begin with an experiment to test whether on an inclined plane with a constant slope, the acceleration seems to be constant regardless of velocity or position on the slope.

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The equations of uniformly accelerated motion

We begin with the return to the situation in which we know `ds, a and v0. Recall that the two most fundamental equations of motion are

vf = v0 + a `dt,

which was derived by considering what could be concluded from knowledge of v0, a and `dt, and which indicates the by-now obvious fact that when we add the change in velocity `dv = a `dt to the initial velocity we get the final velocity, and

`ds = (vf + v0) / 2 * `dt,

derived from the initially known quantities v0, vf and `dt, which indicates that to get the displacement we multiply the average velocity by the time interval.

Would have also seen that knowing v0, a and `dt, are natural line of reasoning leads to the third fundamental equation

`ds = v0 `dt + .5 a `dt^2.

This equation could have been obtained from the first two by substituting the right-hand side of the first equation for vf in the second and doing the straightforward algebraic simplification.

We note that the two most fundamental equations have the three variables vf, v0 and `dt in common. We've just observed that by eliminating vf we get the third fundamental equation of motion. We can also see that this equation doesn't help in the present situation, where we want to draw some kind of a conclusion from knowledge of `ds, a and v0. None of the three equations we have so far contains more than two of these variables among its four variables, and hence none can be solved for one of the remaining variables.

If we eliminate `dt from the first two equations we can see the we will end up with an equation involving vf, v0, a and `ds, which contains three of the known variables and will therefore be useful in the present situation.

The equation that we get when we eliminate `dt is `ds = ( vf^2 - v0^2 ) / (2a), which we rearrange into the form

vf^2 = v0^2 + 2 a `ds.

This is the fourth fundamental equation of motion.

We observe that we now have four equations, each with four variables, with the property that given any three of the variables v0, vf, `ds, `dt and a, at least one of these equations will contain the three and can hence be solved for the fourth.

These equations therefore permit us to determine from any three of the variables the values of the other two, and hence to solve any problem involving uniformly accelerated motion given sufficient information.

These equations all arise from the first two; these two we understand at the intuitive level as expressing the definition of acceleration and average velocity.

Given displacement, acceleration and initial velocity:  Eliminating vf from the first two equations of motion yields the third; eliminating time interval from the first two equations yields the fourth.  Either can be applied to the given situation.

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The details of the process of eliminating `dt from the two most fundamental equations are shown below.

Details of eliminating time interval from first two equations:  Solve second equation for time interval and substitute into the second.

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Accelerating down an incline through a given distance vs. accelerating for a given time

We now pose the following question:

If a cart coasts a specific distance down a certain incline starting from rest, it will experience a certain velocity change.

Most students will agree that this argument seems persuasive, and that since the conditions seee to be the same, the velocity change will indeed be the same in both cases.

Before you read further you should carefully consider both of these arguments and decide for yourself which seems to be more persuasive.

The equations of motion settle the argument.

The equation vf = v0 + a `dt tells us clearly that if we have the same acceleration over a shorter time interval, our velocity will increase by less. The change in velocity is without a doubt a `dt.

The equation vf^2 = v0^2 + 2 a `ds, on the other hand, tells us that when a and `ds are identical, the difference in the squared velocities will be identical:

Solving for a `dt and a `ds, we obtain the equations

a `dt = vf - v0

a `ds = (vf^2 - v0^2) / 2,

which will both have profound implications later when we study momentum and energy.

Contrasting product of acceleration and time interval with product of acceleration and displacement; one result is proportional to change in v while the other is proportional to the change in the square of v.  Very important distinction.

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