let ( s(t)=8 t^{3}-72 t^{2}+120 t ) be the equation of motion for a particle. find a function for the…

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\neq0), 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)