Question

let ( s(t)=4 t^{3}-12 t^{2}-96 t ) be the equation of motion for a particle. find a function for the velocity.( v(t)= )
where does the velocity equal zero? hint: factor out the gcf.( t= ) and ( t= )
find a function for the acceleration of the particle.( a(t)= )
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Explanation:

Step1: Find the velocity function

The velocity \(v(t)\) is the derivative of the position function \(s(t)\).
Using the power rule \(\frac{d}{dt}(t^n)=nt^{n - 1}\), for \(s(t)=4t^{3}-12t^{2}-96t\), we have:
\(v(t)=\frac{d}{dt}(4t^{3}-12t^{2}-96t)=4\times3t^{2}-12\times2t - 96\)
\(v(t)=12t^{2}-24t - 96\)

Step2: Find when \(v(t) = 0\)

Factor out the GCF (which is \(12\)) from \(v(t)=12t^{2}-24t - 96\).
\(v(t)=12(t^{2}-2t - 8)\)
Factor the quadratic \(t^{2}-2t - 8=(t - 4)(t+ 2)\)
Set \(v(t)=0\), so \(12(t - 4)(t + 2)=0\)
Using the zero - product property \(t-4=0\) or \(t + 2=0\)
\(t = 4\) or \(t=-2\)

Step3: Find the acceleration function

The acceleration \(a(t)\) is the derivative of the velocity function \(v(t)\).
Since \(v(t)=12t^{2}-24t - 96\), using the power rule \(\frac{d}{dt}(t^n)=nt^{n - 1}\)
\(a(t)=\frac{d}{dt}(12t^{2}-24t - 96)=12\times2t-24\)
\(a(t)=24t-24\)

Answer:

\(v(t)=12t^{2}-24t - 96\)
\(t=-2\) and \(t = 4\)
\(a(t)=24t - 24\)