let ( s(t)=t^{3}-21 t^{2}+135 t + 5 ) for ( 0 leq t leq 12 ) denote the position of an object moving along a…

let ( s(t)=t^{3}-21 t^{2}+135 t + 5 ) for ( 0 leq t leq 12 ) denote the position of an object moving along a line. find\n- the velocity at time ( t ) is\n- the acceleration at time ( t ) is\n- find the initial position and the ending position\n- find the total distance traveled by the object\n- find where the velocity is positive use interval notation.\n- find where the acceleration is positive use interval notation.\nif the answer includes more than one interval write the intervals separated by the \union\ symbol, u. if needed enter ( infty ) as inf and ( -infty ) as -inf.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.

let ( s(t)=t^{3}-21 t^{2}+135 t + 5 ) for ( 0 leq t leq 12 ) denote the position of an object moving along a line. find\n- the velocity at time ( t ) is\n- the acceleration at time ( t ) is\n- find the initial position and the ending position\n- find the total distance traveled by the object\n- find where the velocity is positive use interval notation.\n- find where the acceleration is positive use interval notation.\nif the answer includes more than one interval write the intervals separated by the \union\ symbol, u. if needed enter ( infty ) as inf and ( -infty ) as -inf.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.

Answer

Explanation:

Step1: Find velocity

Velocity (v(t)) is the derivative of position function (s(t)). Using power rule ((x^n)^\prime = nx^{n - 1}), for (s(t)=t^{3}-21t^{2}+135t + 5), we have (v(t)=s^\prime(t)=3t^{2}-42t + 135).

Step2: Find acceleration

Acceleration (a(t)) is the derivative of velocity function (v(t)). Differentiating (v(t)=3t^{2}-42t + 135) using power rule, (a(t)=v^\prime(t)=6t-42).

Step3: Find initial and ending positions

Initial position: when (t = 0), (s(0)=0^{3}-21\times0^{2}+135\times0 + 5=5). Ending position: when (t = 12), (s(12)=12^{3}-21\times12^{2}+135\times12 + 5=1728-3024 + 1620+5=329).

Step4: Find where velocity is positive

Set (v(t)=3t^{2}-42t + 135>0). Factor (v(t)=3(t^{2}-14t + 45)=3(t - 5)(t - 9)>0). The roots of (y=(t - 5)(t - 9)) are (t = 5) and (t = 9). Using test - points: For (t<5) (e.g., (t = 0)), (v(0)=3\times(0 - 5)(0 - 9)=135>0). For (5<t<9) (e.g., (t = 6)), (v(6)=3\times(6 - 5)(6 - 9)=-9<0). For (t>9) (e.g., (t = 10)), (v(10)=3\times(10 - 5)(10 - 9)=15>0). Since (0\leq t\leq12), the velocity is positive on ([0,5)\cup(9,12]).

Step5: Find where acceleration is positive

Set (a(t)=6t-42>0). Solve (6t-42>0), (6t>42), (t > 7). Since (0\leq t\leq12), the acceleration is positive on ((7,12]).

Step6: Find total distance traveled

First, find critical points of (v(t)) (where (v(t)=0)): (3t^{2}-42t + 135 = 0), (t^{2}-14t + 45=0), ((t - 5)(t - 9)=0), (t = 5) and (t = 9). (s(0)=5), (s(5)=5^{3}-21\times5^{2}+135\times5 + 5=125-525+675 + 5=280), (s(9)=9^{3}-21\times9^{2}+135\times9 + 5=729-1701+1215+5=248), (s(12)=329). Distance (d=\vert s(5)-s(0)\vert+\vert s(9)-s(5)\vert+\vert s(12)-s(9)\vert) (d=\vert280 - 5\vert+\vert248 - 280\vert+\vert329 - 248\vert) (d = 275+32+81=388).

Answer:

  • The velocity at time (t) is (3t^{2}-42t + 135).
  • The acceleration at time (t) is (6t-42).
  • The initial position is (5) and the ending position is (329).
  • The total distance traveled by the object is (388).
  • The velocity is positive on ([0,5)\cup(9,12]).
  • The acceleration is positive on ((7,12]).