Answer :
Answer:
(a) [tex]v(t) = 6t^2 - 6t - 12[/tex], [tex]a(t) = 12t - 6[/tex]
(b) When [tex]0 \leq t < 0.5[/tex], object is slowing down, when [tex] t > 0.5 [/tex] object is speeding up.
Explanation:
(a) To get the velocity function, we need to take the derivative of the position function.
[tex]v(t) = \frac{ds(t)}{dt} = (2t^{3})^{'} - (3t^{2})^{'} - (12t)^{'} + 6^{'} = 6t^{2} - 6t - 12[/tex]
To get the acceleration function, we need to take the derivative of the velocity function.
[tex]a(t) = \frac{dv(t)}{dt} = (6t^{2})^{'} - (6t)^{'} - (12)^{'} = 12t - 6[/tex]
(b) The object is slowing down when velocity is decreasing by time (decelerating) hence a < 0
[tex]12t - 6 < 0 \\12t < 6 \\t < 0.5[/tex]
On the other hand, object is speeding up when a > 0
[tex]12t - 6 > 0 \\12t > 6 \\t > 0.5[/tex]
Therefore, when [tex] 0 \leq t < 0.5[/tex], object is slowing down, when [tex] t > 0.5 [/tex] object is speeding up.