Interpretation of the Derivative — Question 3

Question 3 - Solution

Given: $$s(t) = 3t^3 - 15t^2 + 18t$$

(a) Velocity and Acceleration:

Velocity is the derivative of position: $$v(t) = s'(t) = 9t^2 - 30t + 18$$

Acceleration is the derivative of velocity: $$a(t) = v'(t) = 18t - 30$$

(b) Particle at Rest:

Set velocity to zero: $$9t^2 - 30t + 18 = 0 \Rightarrow t = \frac{30 \pm \sqrt{(-30)^2 - 4(9)(18)}}{2(9)} = \frac{30 \pm \sqrt{900 - 648}}{18} = \frac{30 \pm \sqrt{252}}{18}$$ $$\sqrt{252} = 6\sqrt{7}, \quad t = \frac{30 \pm 6\sqrt{7}}{18} = \frac{5 \pm \sqrt{7}}{3}$$

So the particle is at rest at: $$\boxed{t = \frac{5 \pm \sqrt{7}}{3}}$$

(c) Particle Moving Forward:

Forward motion means $v(t) > 0$. Use sign chart for: $$v(t) = 9t^2 - 30t + 18 = 9(t - t_1)(t - t_2), \quad \text{where } t_1 = \frac{5 - \sqrt{7}}{3}, \, t_2 = \frac{5 + \sqrt{7}}{3}$$

This quadratic opens upward (coefficient of $t^2$ is positive), so: $$v(t) > 0 \quad \text{on } (0, t_1) \cup (t_2, \infty)$$

So the particle moves forward on: $$\boxed{\left(0, \frac{5 - \sqrt{7}}{3}\right) \cup \left(\frac{5 + \sqrt{7}}{3}, \infty\right)}$$

(d) Acceleration Zero:

$$a(t) = 18t - 30 = 0 \Rightarrow t = \frac{30}{18} = \frac{5}{3}$$

This is when the particle changes concavity or when velocity changes most rapidly.

Conclusion: $$\boxed{a(t) = 0 \text{ when } t = \frac{5}{3} \text{ seconds}}$$