Interpretation of the Derivative — Question 1

Question 1 - Solution

We are given the position function: $$s(t) = t^3 - 6t^2 + 9t$$

(a) Velocity and Acceleration:

Velocity is the first derivative of position: $$v(t) = s'(t) = 3t^2 - 12t + 9$$

Acceleration is the derivative of velocity: $$a(t) = v'(t) = 6t - 12$$

(b) Particle at Rest:

The particle is at rest when $v(t) = 0$: $$3t^2 - 12t + 9 = 0 \Rightarrow t^2 - 4t + 3 = 0 \Rightarrow (t - 1)(t - 3) = 0 \Rightarrow t = 1, 3$$

So, the particle is at rest at $t = \boxed{1}$ and $t = \boxed{3}$.

(c) Speeding Up / Slowing Down:

Speed increases when velocity and acceleration have the same sign; slows down when signs differ.

Critical points from previous steps: $$v(t) = 3t^2 - 12t + 9, \quad a(t) = 6t - 12$$

Velocity is 0 at $t = 1, 3$, acceleration is 0 at $t = 2$

Test signs in intervals:

$t < 1$: $v(t) > 0$, $a(t) < 0$ → slowing down

$1 < t < 2$: $v(t) < 0$, $a(t) < 0$ → speeding up

$2 < t < 3$: $v(t) < 0$, $a(t) > 0$ → slowing down

$t > 3$: $v(t) > 0$, $a(t) > 0$ → speeding up

Conclusion:

• Speeding up: $t \in (1, 2) \cup (3, \infty)$

• Slowing down: $t \in (0, 1) \cup (2, 3)$