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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Position 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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1(ft/sec)/(sec) or ft/sec^2 are the same units the second is the more compact way to write it

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1`(ft/sec)/(sec)` is a way to show how the speed is changing over time

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1each 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)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0\(\frac{(\frac{a}{b})}{b}=\frac{a}{b^2}\)
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