Question

the position function of a particle is given by
$s = t^3 - 1.5t^2 - 28t, t \geq 0.$
(a) when does the particle reach a velocity of 8 m/s?
\boxed{} s

(b) when is the acceleration 0?
\boxed{} s

Explanation:

Step1: Find velocity function

Velocity \(v(t)\) is derivative of \(s(t)\):
\(v(t) = s'(t) = 3t^2 - 3t - 28\)

Step2: Solve \(v(t)=8\) for \(t≥0\)

Set \(3t^2 - 3t - 28 = 8\):
\(3t^2 - 3t - 36 = 0\) → \(t^2 - t - 12 = 0\)
Factor: \((t-4)(t+3)=0\) → \(t=4\) (since \(t≥0\))

Step3: Find acceleration function

Acceleration \(a(t)\) is derivative of \(v(t)\):
\(a(t) = v'(t) = 6t - 3\)

Step4: Solve \(a(t)=0\) for \(t≥0\)

Set \(6t - 3 = 0\):
\(t = \frac{3}{6} = 0.5\)

Answer:

(a) 4
(b) 0.5