- #1

JohnnyGui

- 796

- 51

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:

**in which the**

*at + v*_{0}= v**is the starting velocity**

*v*_{0}2. Now, to calculate the distance, one wouldn't obviously be able to just multiply the given

**by**

*v***since that would consider as if the object has been traveling a constant velocity all along.**

*t*In reality one would have to calculate

**in which**

*(v*_{0}+ a) + (v_{0}+ 2a) + (v_{0}+ 3a) + (v_{0}+ na)**would be the time duration in steps of Δt.**

*n*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

**and**

*v*_{0}**that the object has after a time duration**

*v*(**) and multiply that average velocity by the time**

*t***.**Thus, the formula would be

**which after simplifying gives**

*((at + v*_{0}) + v_{0}) / 2) × t = d

*0.5at*^{2}+*v*t = d_{0}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

**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.**

*0.5at*^{2}+ v_{0}t = dFor example: If an object with a start velocity of 6 m/s accelerates in

*discrete*steps of 3m/s

^{2}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

**would give a traveled distance of 48m.**

*0.5at*^{2}+ v_{0}t = dI thought that the known formula

**should always give accurate results regarding acceleration that increases velocity in**

*0.5at*^{2}+ v_{0}t = d*discrete*steps. Perhaps I'm missing something obvious here?