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

• JohnnyGui
In summary, the conversation discussed the relationship between distance, velocity, and acceleration for an object with discrete changes in velocity due to acceleration. It was determined that the formula 0.5at^2 + v0t = d is accurate for constant acceleration, but may not apply for varying acceleration. The concept of taking the average velocity over a time interval was also discussed, and it was noted that using smaller time intervals can lead to more accurate results. The use of the area under the curve in a velocity-time diagram was also mentioned as a way to understand this concept.
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?

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.

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.

CWatters said:
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?

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.

JohnnyGui said:
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.

Last edited:
rcgldr said:
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!

## 1. What is the acceleration formula?

The acceleration formula, also known as the kinematic equation, is a mathematical expression that relates an object's acceleration, initial velocity, final velocity, and displacement. It is represented as a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

## 2. How is the acceleration formula derived?

The acceleration formula is derived from the basic principles of calculus and the definition of acceleration as the rate of change of an object's velocity. By integrating the equation for velocity, v = v0 + at, with respect to time, we can obtain the acceleration formula, a = (vf - vi)/t.

## 3. Can the acceleration formula be used for objects with changing acceleration?

Yes, the acceleration formula can be used for objects with changing acceleration as long as the acceleration is constant over the given time interval. If the acceleration is not constant, the formula can still be used by dividing the time interval into smaller intervals where the acceleration is constant, and then summing up the individual accelerations.

## 4. What are the units of measurement for the acceleration formula?

The units of measurement for the acceleration formula depend on the units used for acceleration, velocity, and time. In the International System of Units (SI), acceleration is measured in meters per second squared (m/s^2), velocity is measured in meters per second (m/s), and time is measured in seconds (s). Therefore, the units for the acceleration formula are m/s^2 = (m/s - m/s)/s.

## 5. How is the acceleration formula used in real-world applications?

The acceleration formula is used in various real-world applications, such as calculating the acceleration of a car, determining the force needed to lift an object, and predicting the motion of a projectile. It is also used in fields such as engineering, physics, and astronomy to analyze and predict the motion of objects in different scenarios.

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