let ( s(t)=4 t^{3}-12 t^{2}-96 t ) be the equation of motion for a particle. find a function for the…

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)= )\nwhere does the velocity equal zero? hint: factor out the gcf.( t= ) and ( t= )\nfind a function for the acceleration of the particle.( a(t)= )\nquestion help: video message instructor\nsubmit question jump to answer

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)= )\nwhere does the velocity equal zero? hint: factor out the gcf.( t= ) and ( t= )\nfind a function for the acceleration of the particle.( a(t)= )\nquestion help: video message instructor\nsubmit question jump to answer

Answer

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)