Acceleration formula derivation (newtonian physics) [closed]

Acceleration has a simple equation in physics:

$$a = \frac{(v-u)}{t}$$

(should be $$\Delta v / \Delta t$$)

Both $$v$$ and $$u$$ have $$s/t$$ as their formula. So, dividing twice by time gives time squared, so we have that acceleration is $$s/t^2$$.

Now, it is also easy to get that $$v = u+at$$ from the formula. Then, it is graphically derived that

$$s=ut+\frac{1}{2}(v-u)t=ut+\frac{1}{2}at^2$$

as the amount of increase of space due to the $$\Delta\,v-u$$ is a square triangle that has as cathetuses time and the $$\Delta$$ itself in terms of velocity, over a rectangle that is $$u\cdot t$$.

While I understand the graphic part, I don't get what happens in the mathematics:

$$v=u+at$$

then

$$\frac{s}{t}=u+at$$

so we would get

$$s=ut+at^2$$

$$s=ut+\frac{1}{2}at^2$$

I know I am wrong in my math or in the fact that acceleration is measured in $$s/t^2$$ so there is some hidden square value I am missing.. but where?

closed as unclear what you're asking by ZeroTheHero, Buzz, Jon Custer, Kyle Kanos, Bill NJan 25 at 1:32

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• The title is ambiguous, and doesn't tend to match the content of the post. AMdP, your ending formula can be derived from the kinematic equations. Do you need to see the derivation? – David White Jan 24 at 1:55
• Your geometric derivation is fine, you should let that guide your algebra. FWIW, in The Principia, Newton gave geometric proofs for every result. – PM 2Ring Jan 24 at 4:57
• David, thanks, would you suggest a better title? – AMDP Jan 25 at 0:55
• PM 2Ring, what do you mean with "guide"? – AMDP Jan 25 at 0:56
• David thanks again, I don't understand clearly how I can derive the formula so yes, maybe it could be useful to see the correct derivation. – AMDP Jan 25 at 0:59

In your derivation, you made the assumption that $$v = s/t$$. This isn't true when velocities are changing.
Example: Suppose you've traveled 100 meters in 5 seconds. What velocity did you travel at? $$v = s/t$$, predicts that you traveled at $$20m/s$$ for 5 seconds. But that's incorrect. You could have traveled at $$100m/s$$ for a second and then $$0 m/s$$ for 4 seconds. Based only on this information, strictly speaking, there isn't enough information to tell.