Question

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.)

Answer 1

Answer:

(a) $v(t) = 6t^2 - 6t - 12$, $a(t) = 12t - 6$

(b) When $0 \leq t < 0.5$, object is slowing down, when $t > 0.5$ object is speeding up.

Explanation:

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

$v(t) = \frac{ds(t)}{dt} = (2t^{3})^{'} - (3t^{2})^{'} - (12t)^{'} + 6^{'} = 6t^{2} - 6t - 12$

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

$a(t) = \frac{dv(t)}{dt} = (6t^{2})^{'} - (6t)^{'} - (12)^{'} = 12t - 6$

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

$12t - 6 < 0 \\12t < 6 \\t < 0.5$

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

$12t - 6 > 0 \\12t > 6 \\t > 0.5$

Therefore, when $0 \leq t < 0.5$, object is slowing down, when $t > 0.5$ object is speeding up.