please show steps: The position of a particle moving along an x-axis is given by

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please show steps:

The position of a particle moving along an x-axis is given by x = 12.0t2 - 6.00t3, where x is in meters and t is in seconds. Determine the position, the velocity, and the acceleration of the particle at t = 4.00s. what is the maximum positive coordinates reached by the particle and at what time is it reached? What is the maximum positive velocity reached by the particle and at what time is it reached? What is the acceleration of the particle at that instant the particle is not moving (other than at t = 0?) Determine the average velocity of the particle between t = 0 and t = 4.00s. Number Units Number Units Number Units Number Units Number Units Number Units Number Units Number Units Number Units

Explanation / Answer

(a) x=12t^2-6t^3 t=4

x= 12(4)^2-6(4)^3=-192m

(b) diffrentiating x w.r.t t we get dx/dt=v=24t-18t^2

v(4)=24(4)-18(4)^2

v(4)=-192 m/s

(c) acceleration=dv/dt=24-36t

a(4)=24-36(4)=-120 m/s^2

(d) dx/dt=0

24t-18t^2=0

24-18t=0

t=1.33 s

t>1.33s v<0

t<1.33s v>0

max x coordinate is reached when t=1.33s

x=12(1.33)^2-6(1.33)^3=7.11m

(e) time=1.33s

(F) dv/dt=0

24=36t

t=0.66 s

maximum positive velocity=v(0.66)=24(0.66)-18(0.66)^2=7.84 m/s

(g) time=0.66s

(h)acceleration at t=1.33s=24-36(1.33)=-23.88 m/s^2

(i) average velocity=total dispalcement/total time

x(4)=-192m

x(0)=0m

avg velocity= -192/4=-48 m/s

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