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?