# **Background and Explanation** This problem involves kinematics in one dimension, where position $S(t)$, velocity $V(t)$, and acceleration $a(t)$ are related through differentiation. Velocity is the rate of change of position, and acceleration is the rate of change of velocity. Speed increases when velocity and acceleration have the same sign (both positive or both negative), while distance increases when velocity is positive. --- # **Solution** First, we establish the velocity and acceleration functions by differentiating the position function: $$ \begin{aligned} S &= 2t^3 - 9t^2 + 12t - 4 \\ V &= \frac{dS}{dt} \\ &= \frac{d}{dt}(2t^3 - 9t^2 + 12t - 4) \\ &= 6t^2 - 18t + 12 \\ a &= \frac{dV}{dt} \\ &= \frac{d}{dt}(6t^2 - 18t + 12) \\ a &= 12t - 18 \end{aligned} $$ ## Part (i): Time intervals where distance is increasing Distance $S$ is increasing when velocity is positive: $$\frac{dS}{dt} > 0$$ So: $$ \begin{aligned} 6t^2 - 18t + 12 &> 0 \\ 6(t^2 - 3t + 2) &> 0 \\ t^2 - 2t - t + 2 &> \frac{0}{6} \\ t(t-2) - 1(t-2) &> 0 \\ (t-2)(t-1) &> 0 \\ t-1 &> 0 \\ t &> 1 \\ &\text{or } t-2 > 0 \end{aligned} $$ $\Rightarrow t \in (2, \infty)$ Either $t < 1$ or $t < 2$ $\Rightarrow t \in [0, 1) \quad$ as $t > 0$ So $t \in [0, 1) \cup (2, \infty)$ ## Part (ii): Time intervals where velocity is increasing Velocity is increasing when acceleration is positive: $$\frac{dV}{dt} > 0 \Rightarrow 12t - 18 > 0$$ $$t > \frac{18}{12} \Rightarrow t > \frac{3}{2}$$ So $t \in \left(\frac{3}{2}, \infty\right)$ ## Part (iii): Time intervals where speed is increasing If $V(t) > 0$ and $a(t) > 0$, speed increases. If $V(t) < 0$ and $a(t) < 0$, speed also increases. From previous result $t$ changes sign at $t = \mathbf{1}, 2$. $V(t) > 0$ for $t < 1$ and $t > 2$ Also $V(t) < 0$ for $1 < t < 2$ and $a(t) > 0$ for $t > 1.5$. Also $a(t) < 0$ for $t < 1.5$ Checking intervals: | Interval | Signs | Speed Behavior | | :--- | :--- | :--- | | At $(0, 1)$ | $V > 0$ ; $a < 0$ | speed decreasing | | At $(1, 1.5)$ | $V < 0$ ; $a < 0$ | so speed increasing | | At $(1.5, 2)$ | $V < 0$ ; $a > 0$ | so speed decreasing | | At $(2, \infty)$ | $V > 0$ ; $a > 0$ | speed increasing | So $t \in (1, 1.5) \cup (2, \infty)$ ## Part (iv): Speed at $t = \frac{3}{2}$ At $t = \frac{3}{2}$: $$ \begin{aligned} V &= 6\left(\frac{3}{2}\right)^2 - 18\left(\frac{3}{2}\right) + 12 \\ &= \frac{54}{4} - \frac{54}{5} + 12 \\ &= 12 \text{ m/s} \end{aligned} $$ ## Part (v): Total distance travelled in $[0, 4]$ $v(t) = 6t^2 - 18t + 12$ has roots at $t = 1$ and $t = 2$. Calculate position at critical points: $$ \begin{aligned} S(0) &= 2(0)^3 - 9(0)^2 + 12(0) - 4 = -4 \\ S(1) &= 2(1)^3 - 9(1)^2 + 12(1) - 4 = 1 \\ S(2) &= 2(2)^3 - 9(2)^2 + 12(2) - 4 = 0 \\ S(4) &= 2(4)^3 - 9(4)^2 + 12(4) - 4 = 36 \end{aligned} $$ Total distance $= |1 - (-4)| + |0 - 1| + |36 - 0|$ $$= 4 + 1 + 36 = 41 \text{ m}$$ --- # **Key Formulas or Methods Used** - $V = \frac{dS}{dt}$ (velocity as derivative of position) - $a = \frac{dV}{dt} = \frac{d^2S}{dt^2}$ (acceleration as derivative of velocity) - Distance increasing when $V > 0$ - Velocity increasing when $a > 0$ - Speed increasing when $V$ and $a$ have the same sign ($V \cdot a > 0$) - Total distance travelled $= \sum |S(t_{i}) - S(t_{i-1})|$ over intervals where velocity changes sign --- # **Summary of Steps** 1. **Find derivatives**: Differentiate $S(t)$ to get $V(t)$, then differentiate $V(t)$ to get $a(t)$. 2. **Part (i)**: Solve $V(t) > 0$ by factoring the quadratic to find when distance increases. 3. **Part (ii)**: Solve $a(t) > 0$ to find when velocity increases. 4. **Part (iii)**: Determine sign changes of $V(t)$ (at $t=1, 2$) and $a(t)$ (at $t=1.5$), then test intervals to find where $V$ and $a$ have the same sign. 5. **Part (iv)**: Substitute $t = \frac{3}{2}$ into $V(t)$ and calculate the magnitude. 6. **Part (v)**: Evaluate $S(t)$ at endpoints and at times where $V(t) = 0$ (direction changes), then sum the absolute differences between consecutive positions.