Question

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

Answer 1

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

Answer 2

(ft/sec)/(sec) or ft/sec^2 are the same units

the second is the more compact way to write it

Answer 3

`(ft/sec)/(sec)` is a way to show how the speed is changing over time

Answer 4

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)

Answer 5

\(\frac{(\frac{a}{b})}{b}=\frac{a}{b^2}\)