Let st 3t4 8t3 192t for 0 t 6 denote the position of an o

Let s(t) = 3t^4 + 8t^3 - 192t for 0 t 6 denote the position of an object moving along a line. Find The velocity at time t is The acceleration at time t is Find the initial position and the ending position Find the total distance traveled by the object Find where the velocity is positive Use interval notation. If the answer includes more than one interval write the intervals separated by the \"union\" symbol, U. If needed enter infinity as INF and -infinity as -INF.

Solution

Position of the object s(t) = 3t ⁴ +8t ³ -192 t            ------------( 1)

Velocity of the object v(t) = ds(t) / dt

                                     = 3( 4t ³) +8(3t ²) -192

                                     = 12t ³ +24 t ² -192

Acceleration of the object a(t) = dv(t) / dt

                                            = 12( 3t ²) + 24 (2t)

                                            = 36 t ² + 48 t

For initial position you put t = 0 in equation( 1),

So, initial position s(0) = 3(0 ⁴) +8(0 ³) -192 (0)

                                 = 0

For ending position you put t = 6 in equation( 1),

Ending position s(6) = 3(6 ⁴) +8(6 ³) -192 (6)

                              = 3(1296)+8(216)-1152

                              = 4464

Total distance travelled = s(6) -s(0)

                                  =4464 - 0

                                  = 4464