Question

let ( s(t)=4 t^{3}+6 t^{2}-144 t ) be the equation of motion for a particle. find a function for ( 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)= ) question help: video message instructor

Explanation:

Step1: Find the velocity function \(v(t)\)

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 \(s(t)=4t^{3}+6t^{2}-144t\), we have:
\(v(t)=\frac{d}{dt}(4t^{3}+6t^{2}-144t)=4\times3t^{2}+6\times2t-144\)
\(v(t)=12t^{2}+12t - 144\)

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

Set \(v(t)=0\), so \(12t^{2}+12t - 144 = 0\). Factor out the GCF (which is \(12\)):
\(12(t^{2}+t - 12)=0\). Then solve \(t^{2}+t - 12=0\).
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b = 1\), \(c=-12\)), or factoring \(t^{2}+t - 12=(t + 4)(t - 3)=0\).
\(t=-4\) or \(t = 3\). Since \(t\geq0\) in the context of motion (usually time starts from \(t = 0\)), we consider \(t = 3\) (if we assume \(t\geq0\)). But if we just solve the quadratic equation without the physical - context assumption \(t^{2}+t - 12=0\) gives \(t=-4\) and \(t = 3\)

Step3: Find the acceleration function \(a(t)\)

The acceleration function \(a(t)\) is the derivative of the velocity function \(v(t)\).
Since \(v(t)=12t^{2}+12t - 144\), using the power rule \(\frac{d}{dt}(t^n)=nt^{n - 1}\)
\(a(t)=\frac{d}{dt}(12t^{2}+12t - 144)=12\times2t+12\)
\(a(t)=24t + 12\)

Answer:

\(v(t)=12t^{2}+12t - 144\); \(t=-4\) and \(t = 3\); \(a(t)=24t + 12\)