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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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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

  2. jim_thompson5910
    • one year ago
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    (ft/sec)/(sec) or ft/sec^2 are the same units the second is the more compact way to write it

  3. jim_thompson5910
    • one year ago
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    `(ft/sec)/(sec)` is a way to show how the speed is changing over time

  4. jim_thompson5910
    • one year ago
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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)

  5. zzr0ck3r
    • one year ago
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    \(\frac{(\frac{a}{b})}{b}=\frac{a}{b^2}\)

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