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.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.