Question

let ( s(t)=8 t^{3}-72 t^{2}+120 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 function \(v(t)\) is the derivative of the position function \(s(t)\).
Using the power rule \(\frac{d}{dt}(t^n)=nt^{n - 1}\), for \(s(t)=8t^{3}-72t^{2}+120t\), we have:
\(v(t)=\frac{d}{dt}(8t^{3}-72t^{2}+120t)\)
\(v(t)=8\times3t^{2}-72\times2t + 120\times1\)
\(v(t)=24t^{2}-144t + 120\)

Step2: Find when the velocity is zero

Set \(v(t)=0\), so \(24t^{2}-144t + 120 = 0\).
Factor out the GCF (which is \(24\)): \(24(t^{2}-6t + 5)=0\).
Since \(24
eq0\), we solve \(t^{2}-6t + 5=0\).
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(at^{2}+bt + c=t^{2}-6t + 5\) (\(a = 1\), \(b=-6\), \(c = 5\)) or factoring \(t^{2}-6t + 5=(t - 1)(t - 5)=0\).
So \(t=1\) and \(t = 5\)

Step3: Find the acceleration function

The acceleration function \(a(t)\) is the derivative of the velocity function \(v(t)\).
Since \(v(t)=24t^{2}-144t + 120\), using the power rule \(\frac{d}{dt}(t^n)=nt^{n - 1}\)
\(a(t)=\frac{d}{dt}(24t^{2}-144t + 120)\)
\(a(t)=24\times2t-144\times1+0\)
\(a(t)=48t-144\)

Answer:

\(v(t)=24t^{2}-144t + 120\)
\(t = 1\) and \(t = 5\)
\(a(t)=48t-144\)