a particle moves along a straight line and its position at time ( t ) is given by ( s(t)=2 t^{3}-21 t^{2}+36…

a particle moves along a straight line and its position at time ( t ) is given by ( s(t)=2 t^{3}-21 t^{2}+36 t ) where ( s ) is measured in feet and ( t ) in seconds.\nfind the velocity (in ( mathrm{ft} / mathrm{sec} ) ) of the particle at time ( t = 0 ):\nthe particle stops moving (i.e. is in a rest) twice,\nfirst when ( t=),\nand again when ( t=)\nwhat is the position of the particle at time 14? ( mathrm{ft} )\nfinally, what is the total distance the particle travels between time 0 and time 14? ( mathrm{ft} )\nquestion help: message instructor

a particle moves along a straight line and its position at time ( t ) is given by ( s(t)=2 t^{3}-21 t^{2}+36 t ) where ( s ) is measured in feet and ( t ) in seconds.\nfind the velocity (in ( mathrm{ft} / mathrm{sec} ) ) of the particle at time ( t = 0 ):\nthe particle stops moving (i.e. is in a rest) twice,\nfirst when ( t=),\nand again when ( t=)\nwhat is the position of the particle at time 14? ( mathrm{ft} )\nfinally, what is the total distance the particle travels between time 0 and time 14? ( mathrm{ft} )\nquestion help: message instructor

Answer

Explanation:

Step1: Find the velocity function

The velocity function (v(t)) is the derivative of the position function (s(t)). Using the power rule ((x^n)^\prime = nx^{n - 1}), if (s(t)=2t^{3}-21t^{2}+36t), then (v(t)=s^\prime(t)=6t^{2}-42t + 36).

Step2: Find the velocity at (t = 0)

Substitute (t = 0) into (v(t)): (v(0)=6(0)^{2}-42(0)+36=36).

Step3: Find when the particle stops moving

Set (v(t)=0), so (6t^{2}-42t + 36 = 0). Divide through by (6): (t^{2}-7t + 6=0). Factor: ((t - 1)(t - 6)=0). Using the zero - product property (t-1 = 0) or (t - 6=0), so (t = 1) or (t = 6).

Step4: Find the position at (t = 14)

Substitute (t = 14) into (s(t)): (s(14)=2(14)^{3}-21(14)^{2}+36(14)) (=2\times2744-21\times196 + 36\times14) (=5488-4116+504) (=1876).

Step5: Find the total distance

We need to consider the intervals ([0,1]), ([1,6]), and ([6,14]).

  • For (t\in[0,1]): (v(t)=6t^{2}-42t + 36), (v(t)>0) (test (t = 0.5), (v(0.5)=6\times(0.5)^{2}-42\times(0.5)+36=1.5-21 + 36=16.5>0)) (s(1)-s(0)=(2\times1^{3}-21\times1^{2}+36\times1)-(2\times0^{3}-21\times0^{2}+36\times0)=2 - 21+36=17)
  • For (t\in[1,6]): (v(t)=6t^{2}-42t + 36), (v(t)<0) (test (t = 2), (v(2)=6\times2^{2}-42\times2+36=24-84 + 36=-24<0)) (s(6)-s(1)=(2\times6^{3}-21\times6^{2}+36\times6)-(17)=(432-756 + 216)-17=-125) (distance is (|s(6)-s(1)| = 125))
  • For (t\in[6,14]): (v(t)=6t^{2}-42t + 36), (v(t)>0) (test (t = 7), (v(7)=6\times7^{2}-42\times7+36=294-294+36=36>0)) (s(14)-s(6)=(1876)-(2\times6^{3}-21\times6^{2}+36\times6)=1876-(432-756 + 216)=1876 + 108=1984)

The total distance (D=17 + 125+1984=2126).

Answer:

The velocity at (t = 0) is (36) ft/sec. The particle stops at (t = 1) and (t = 6). The position at (t = 14) is (1876) ft. The total distance is (2126) ft.