Position function to velocity function to acceleration. Will post equation in comment

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Position function to velocity function to acceleration. Will post equation in comment

Calculus1
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Position function P(t) is: \[-16t^2 + V_0t + S_0\] Velocity P'(t) is: \[-32t + V_0 + 0\] Acceleration P"(t) is: -32 + 0 My question is that acceleration if (ft/sec)/(sec) or ft/sec^2 How do we show this in calculus? Using the 1st & 2nd derivative I don't see that acceleration is per seconds squared
(ft/sec)/(sec) or ft/sec^2 are the same units the second is the more compact way to write it
`(ft/sec)/(sec)` is a way to show how the speed is changing over time

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each time you take the derivative, you're seeing how y changes as t changes basically you're computing dy/dt and that's how the position changes to velocity (ft ---> ft/s) and how velocity changes to acceleration (ft/s ---> (ft/s)/s = ft/s^2)
\(\frac{(\frac{a}{b})}{b}=\frac{a}{b^2}\)

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