How can the 0.5 in the acceleration and distance formula be explained?

[JohnnyGui]

Hello all,

First of all I want to let you know that my question is very basic and that it involves discrete changes in velocity due to acceleration for every given Δt. I was trying to derive the relationship between the distance and acceleration in a formula and here's what I came up with:

1. I was able to conclude that to calculate the velocity after a time t in which discrete acceleration is involved the formula would be: at + v₀ = v in which the v₀ is the starting velocity

2. Now, to calculate the distance, one wouldn't obviously be able to just multiply the given v by t since that would consider as if the object has been traveling a constant velocity all along.
In reality one would have to calculate (v₀ + a) + (v₀ + 2a) + (v₀ + 3a) + (v₀ + na) in which n would be the time duration in steps of Δt.

3. However, to give an approximation of the distance traveled without doing the whole hassle in point 2, one could just take the average velocity of v₀ and v (that the object has after a time duration t) and multiply that average velocity by the time. Thus, the formula would be ((at + v₀) + v₀) / 2) × t = d which after simplifying gives 0.5at² + v₀t = d
Question: Is taking the average the reason why there's a "0.5" in the formula that gives the relationship of acceleration and distance?

However, here's my problem. The formula 0.5at² + v₀t = d doesn't always seem to give correct answers even for a discrete acceleration over time when I compare its results to the results of the formula that I've shown in point 2.
For example: If an object with a start velocity of 6 m/s accelerates in discrete steps of 3m/s² for a time duration of 4 seconds, I'd expect that it would have traveled 6 + 9 + 12 + 15 = 42m at t=4. However, filling the values in the formula 0.5at² + v₀t = d would give a traveled distance of 48m.

I thought that the known formula 0.5at² + v₀t = d should always give accurate results regarding acceleration that increases velocity in discrete steps. Perhaps I'm missing something obvious here?

 

[Jonathan Scott]

The average velocity over an interval is the total distance divided by the total time. If acceleration is constant, the average velocity is half way between the initial velocity and the final velocity for an interval, hence the formula. If it is not constant, the formula does not apply. As a special case, for your case of discrete steps, then if your sudden change in speed occurred at the half way time within each step, instead of at the end of it, the results would be the same as for constant acceleration.

 

[CWatters]

Your equation d = 0.5at² + v₀t is consistent with the SUVAT equations for constant acceleration...

https://en.wikipedia.org/wiki/Equations_of_motion

However it's not clear what you mean by "accelerates in discrete steps". Perhaps plot a graph of velocity vs time. It it's not a straight line then the acceleration isn't constant and the equations don't apply.

 

[JohnnyGui]

Your equation d = 0.5at² + v₀t is consistent with the SUVAT equations for constant acceleration...

https://en.wikipedia.org/wiki/Equations_of_motion

However it's not clear what you mean by "accelerates in discrete steps". Perhaps plot a graph of velocity vs time. It it's not a straight line then the acceleration isn't constant and the equations don't apply.

I knew I was missing something obvious here.

This explains why it also always gives a higher value of distance than with discrete steps of acceleration (discrete being a "sudden" increase in velocity at each fixed Δt) since there's a constant velocity increase even between the Δt.

So if I understand correctly, this formula d = 0.5at² + v₀t is even accurate if there's constant acceleration in infinitesimally small Δt?

 

[Jonathan Scott]

The formula assumes the average speed over the total elapsed time is exactly half way between the initial speed and the final speed, as you noted in your original post. This is always true if the acceleration is constant during the total time. If the acceleration varies, the formula cannot be used.

 

[rcgldr]

If an object with a start velocity of 6 m/s accelerates in discrete steps of 3m/s² for a time duration of 4 seconds, I'd expect that it would have traveled 6 + 9 + 12 + 15 = 42m at t=4.

Assuming acceleration is constant, for a relatively large Δt = 1, you still need to take the average velocity for each step (6+9)/2 + (9+12)/2 + (12+15)/2 + (15+18)/2 = 48.

If you want to use just the starting or ending velocities for each time period, you need to use a smaller Δt. If you use the starting velocities, the result is 48 - 6 Δt. If you use the ending velocities, the result is 48 + 6 Δt. As Δt approaches zero, both methods approach 48.

 

[JohnnyGui]

Assuming acceleration is constant, for a relatively large Δt = 1, you still need to take the average velocity for each step (6+9)/2 + (9+12)/2 + (12+15)/2 + (15+18)/2 = 48.

If you want to use just the starting or ending velocities for each time period, you need to use a smaller Δt. If you use the starting velocities, the result is 48 - 6 Δt. If you use the ending velocities, the result is 48 + 6 Δt. As Δt approaches zero, both methods approach 48.

Thanks, your explanation helped me a lot and I was able to conclude all that by calculating the area beneath a line in a v t diagram with constant acceleration in the ways you mentioned.Thank you all for your help!