Need some feedback on a variation of the [[D = V°(t) + 1/2(A)(t)^2]] formula

In summary, the conversation discusses the formula for finding the distance traveled by an object with constant acceleration and explores the idea of a constantly increasing acceleration and its effect on velocity. The modified formula for finding the velocity in such cases is also discussed. However, the relationship between constantly increasing acceleration and distance traveled is not as straightforward and may require further exploration.
  • #1
realfuzzhead
14
0
Hello guys, hope everything is going well. Anyways, I was thinking recently about the forumla for finding the distance an object has traveled with a constant acceleration using the formula

D = V°(t) + 1/2 (A)(t)2

So basically when thinking about this formula, I like to think about an object that starts from origin at rest (0 m/s), experiencing a constant force that causes a constant acceleration, which increases the velocity linearly, which increases the distance Parabolically (maybe not exactly a parabola but not linear is what I mean.)

Ok so here is an example.

An Object is at rest at Time(T) = 0 seconds.
The object is experiencing a constant force (arbitrary because I wish to not introduce numbers) that causes a constant acceleration.

Let's say the object is accelerating at 2 m/s2
A = 2 m/s2
V°(initial V) = 0
V = V°+ (A)(T)

Then the distance is equal to

[[ D = (0)(t) + 1/2(2)(t)^2 = (1)(t)2 = t2 ]]

So the distance increases as a parabola, t^2.



here is a graph representing these functions according to what we just put down above. Distance = T2, V = 2(t), A = 2.

crmv6.jpg




so once again, a constant acceleration causes a linear velocity which causes a parabolic increase in distance.
___________________________________________________________________________



Here is where it got interesting to me, instead of considering a constant force that causes a constant acceleration, which causes a linearly increasing velocity which creates a parabolic increasing distance, I was wondering what kind of effect a constantly INCREASING force, which causes a constantly INCREASING acceleration, has on velocity?

My idea was that maybe we should just use the orignal equation, (D = V°(t) + 1/2 (A)(t)^2),
and adjust it to help us find the velocity if we know how our acceleration is constantly increasing.



So here is the way I have attempted to translate the formula to fit our new paradigm of a constantly increasing acceleration instead of a constantly increasing velocity.

I basically take every variable except for time, and substitute it's derivative. The derivative of a Distance vs. Time graph is an objects instanteous velocity.

So I replace D, in (D = V°(t) + 1/2 (A)(t)^2), with V, for Velocity.

The deriviate of a velocity vs time graph is an objects acceleration. So I replace V°(initial Velocity) with A°(initial acceleration)

we now have (V = A°(t) + 1/2 (A)(t)^2) [ we substituted V for D, and A° for V°]

The derivative of a acceleration vs time graph is an objects change in acceleration. For the sake of this I am now going to refer to this objects change of acceleration as Ω, this will make it easier than typing out that I am referring to the acceleration of the objects acceleration

We now replace A, in our original problem (D = V°(t) + 1/2 (A)(t)^2), with Ω, which gives us

(V = A°(t) + 1/2 (Ω)(t)^2).

So what does this say? It says for an object that is experiencing a constantly increasing force, which results in a constantly increasing acceleration, we can find the velocity by multiplying the original acceleration by the time, and adding that to the average acceleration times the time squared.

So now let's use those same numbers from earlier, but move them up an order of magnitude if you know what I mean.

I mean what was a constant acceleration, will now become a constant Ω, (increase in acceleration). What was a linear increase in velocity, will now become a linear increase of acceleration. Well by this logic, the Velocity will then take on the role of the distance in the orignal equation.So now the velocity will experience a parabolic type increase.


____________________________________________________________________________
JUST FOR CLARITY

Values of the original graph

Ω(change of acceleration) -- 0

A(acceleration) -- 2 m/s2

V(Velocity) -- (A)*(t) = (2)(t)

D(Distance) = [V°(t) + 1/2(A)(t2)]= --> (0)(t) + 1/2(2)(t2) = ---> t2

_____________________________________________________________________________

Now we are going to move each of the values up one step to the higher magnitude of derivative above it. (notice velocity = f'(x) for D vs. T graphs. Accel. = f'(x) for V vs. T charts etc)

Values of the new graph

Ω(change of acceleration) -- 2 m/s3

A(acceleration) -- (Ω)(t) = --> (2)(t)

V(Velocity) -- [A°(t) + 1/2(Ω)(t2)]= --> (0)(t) + 1/2(2)(t2) = ---> t2

D(Distance) = I don't know what this is.. I don't even want to begin thinking about how an every increasing acceleration effects the distance traveled. Maybe it effects it exponentially?












So once again, for clarity,( maybe over clarity) , a force is causing an object that starts at rest and from an initial acceleration of 0m/s2, to increase it's acceleration at 2 m/s3. The constant increase in accel. causes the acceleration to increase linearly. The linear increase in acceleration causes the velocity to increase Parabolically!

Here is the graph

2hqz5fl.jpg




as you can see, a constantly increasing acceleration effects velocity in the same way that constantly increasing velocity effects the distance traveled?



so is this even important? or is it blatenly obvious? Is there an easier way to come to this or did a make a terrible error?
 
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  • #2


Hi there,

Thank you for sharing your thoughts and ideas on this topic. Your approach to finding the velocity for an object with a constantly increasing acceleration is definitely interesting and worth exploring further.

One thing to keep in mind is that in real-life scenarios, it is rare to have a constantly increasing acceleration. Most of the time, the acceleration may vary or may even become constant over a certain period of time. In those cases, the original formula you mentioned (D = V°(t) + 1/2 (A)(t)^2) would still be applicable.

However, your approach can be useful in cases where the acceleration is constantly increasing, such as in rocket launches or other situations involving a change in force. In these cases, your modified formula (V = A°(t) + 1/2 (Ω)(t)^2) can give us a better understanding of the velocity of the object.

As for the effect of constantly increasing acceleration on distance, it is not as straightforward as the relationship between velocity and distance. This is because the distance traveled is dependent on the velocity of the object at a given time, and the velocity is dependent on the acceleration. So, a constantly increasing acceleration would lead to a constantly increasing velocity, which would result in a constantly increasing distance traveled. However, the relationship between distance and time may not be as simple as a parabola, as it would depend on the specific values of acceleration and time.

Overall, your approach is a valid and interesting way to think about the relationship between velocity and acceleration. It may not be applicable in all scenarios, but it can definitely be useful in certain cases. Keep exploring and questioning, as that is what science is all about!
 

1. What is the purpose of the D = V°(t) + 1/2(A)(t)^2 formula?

The D = V°(t) + 1/2(A)(t)^2 formula is used to calculate the displacement of an object over a given period of time. It takes into account the initial velocity (V°), acceleration (A), and time (t) to determine the distance (D) traveled by the object.

2. Can this formula be applied to any type of motion?

Yes, this formula can be used for any type of motion, whether it is linear, circular, or projectile. As long as the values of initial velocity, acceleration, and time are known, this formula can be used to calculate the displacement.

3. How accurate is this formula?

This formula is highly accurate for calculating displacement as long as the initial velocity and acceleration remain constant throughout the given period of time. It may not provide accurate results for situations where there are significant changes in acceleration or when dealing with non-uniform motion.

4. Can this formula be modified for different units of measurement?

Yes, this formula can be modified to accommodate different units of measurement. For example, if the initial velocity is given in meters per second (m/s) and the acceleration is given in meters per second squared (m/s^2), then the displacement will be calculated in meters (m).

5. Is there a limit to the number of variables that can be included in this formula?

No, there is no limit to the number of variables that can be included in this formula. However, it is important to ensure that all variables are consistent with each other in terms of units of measurement.

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