The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 12t + 6 for t ≥ 0 (a) find the velocity and acceleration functions. v(t) = 6t2−6t−12 a(t) = 12t−6 (b) determine the time intervals when the object is slowing down or speeding up. (enter your answers using interval notation.)

(a) ,

(b) When , object is slowing down, when object is speeding up.

Explanation:

(a) To get the velocity function, we need to take the derivative of the position function.

To get the acceleration function, we need to take the derivative of the velocity function.

(b) The object is slowing down when velocity is decreasing by time (decelerating) hence a < 0

On the other hand, object is speeding up when a > 0

Therefore, when , object is slowing down, when object is speeding up.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y