## anonymous one year ago Position function to velocity function to acceleration. Will post equation in comment

1. anonymous

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

2. jim_thompson5910

(ft/sec)/(sec) or ft/sec^2 are the same units the second is the more compact way to write it

3. jim_thompson5910

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

4. jim_thompson5910

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)

5. zzr0ck3r

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