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)= )…

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)