(3 points)\nlet ( s(t)=t^{2}-2 t - 4 ) for ( 0 leq t leq 8 ) denote the position of an object moving along a…

(3 points)\nlet ( s(t)=t^{2}-2 t - 4 ) for ( 0 leq t leq 8 ) 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.

(3 points)\nlet ( s(t)=t^{2}-2 t - 4 ) for ( 0 leq t leq 8 ) 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:

Velocity

Step1: Differentiate position function

The velocity (v(t)) is the derivative of (s(t)). Using the power rule (\frac{d}{dt}(t^n)=nt^{n - 1}), for (s(t)=t^{2}-2t - 4), we have (v(t)=\frac{d}{dt}(t^{2}-2t - 4)). [v(t)=2t-2]

Acceleration

Step1: Differentiate velocity function

The acceleration (a(t)) is the derivative of (v(t)). Since (v(t)=2t - 2), then (a(t)=\frac{d}{dt}(2t - 2)). [a(t)=2]

Initial and ending position

Step1: Find initial position

The initial position is when (t = 0). Substitute (t = 0) into (s(t)): (s(0)=0^{2}-2\times0 - 4). [s(0)=-4]

Step2: Find ending position

The ending position is when (t = 8). Substitute (t = 8) into (s(t)): (s(8)=8^{2}-2\times8 - 4). [s(8)=64-16 - 4=44]

Total distance traveled

Step1: Find critical points of velocity

Set (v(t)=0), so (2t-2 = 0), which gives (t = 1).

Step2: Evaluate position at critical and end - points

(s(0)=-4), (s(1)=1^{2}-2\times1 - 4=-5), (s(8)=44). The distance from (t = 0) to (t = 1) is (|s(1)-s(0)|=|-5-(-4)| = 1). The distance from (t = 1) to (t = 8) is (|s(8)-s(1)|=|44-(-5)| = 49). The total distance (D=1 + 49=50).

Where velocity is positive

Step1: Solve (v(t)>0)

Set (2t-2>0), add 2 to both sides: (2t>2), divide by 2: (t > 1). Since (0\leq t\leq8), the interval is ((1,8]).

Where acceleration is positive

Step1: Analyze (a(t))

Since (a(t)=2>0) for all (t) in the domain (0\leq t\leq8), the interval is ([0,8]).

Answer:

  • The velocity at time (t) is (2t - 2).
  • The acceleration at time (t) is (2).
  • The initial position is (-4) and the ending position is (44).
  • The total distance traveled by the object is (50).
  • The velocity is positive on the interval ((1,8]).
  • The acceleration is positive on the interval ([0,8]).