(i)

\begin{align} \text{When } & t = 2, \\ s & = 2(2)^2 + (2) \\ & = 10 \text{ m} \end{align}

(ii) Since the particle did not change direction, the displacement after 2 seconds is equal to the distance travelled

$$ \text{Distance travelled} = 10 \text{ m} $$

(i)

\begin{align} \text{When } & t = 0, \\ v & = 5(0) - {1 \over 2}(0)^2 \\ & = 0 \text{ m/s} \end{align}

(ii)

\begin{align} \text{When } & v = 0, \\ 0 & = 5t - {1 \over 2}t^2 \\ 0 & = 10t - t^2 \\ 0 & = t(10 - t) \\ \\ t = 0 \phantom{00} & \text{or} \phantom{000} 10 - t = 0 \\ & \phantom{or00010} - t = -10 \\ & \phantom{or00010()0} t = 10 \end{align}

(i)

\begin{align} \text{When } & a = 0.05, \\ 0.05 & = t - 5t^2 \\ 1 & = 20t - 100t^2 \\ 100t^2 - 20t + 1 & = 0 \\ (10t)^2 - 2(10t)(1) + (1)^2 & = 0 \\ (10t - 1)^2 & = 0 \\ 10t - 1 & = 0 \\ 10t & = 1 \\ t & = {1 \over 10} \\ & = 0.1 \end{align}

(ii) If the particle is decelerating at $6 \text{ m/s}^2$, the particle is accelerating at $-6 \text{ m/s}^2$.

\begin{align} \text{When } & a = -6, \\ -6 & = t - 5t^2 \\ 0 & = t - 5t^2 + 6 \\ 0 & = 5t^2 - t - 6 \\ 0 & = (5t - 6)(t + 1) \\ \\ 5t - 6 = 0 \phantom{000} & \text{or} \phantom{000} t + 1 = 0 \\ 5t = 6 \phantom{000} & \phantom{or000+1} t = -1 \text{ (Reject, } t \ge 0) \\ t = {6 \over 5} \phantom{0()} & \\ t = 1.2 \phantom{0.} & \end{align}

(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} (7t^2 - 10t) \\ & = 7(2)(t) - 10(1) \\ & = 14t - 10 \end{align}

(ii)

\begin{align} v & = 14t - 10 \\ \\ \text{When } & t = 3, \\ v & = 14(3) - 10 \\ & = 32 \text{ m/s} \end{align}

(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} (7t^2 - 2t^3) \\ & = 7(2)(t) - 2(3)(t^2) \\ & = 14t - 6t^2 \end{align}

(ii)

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} (14t - 6t^2) \\ & = 14(1) - 6(2)(t) \\ & = 14 - 12t \end{align}

(iii)

\begin{align} a & = 14 - 12t \\ \\ \text{When } & t = 2, \\ a & = 14 - 12(2) \\ & = -10 \text{ m/s}^2 \end{align}

(i)

\begin{align} v & = \int a \phantom{0} dt \\ & = \int (t - 9) \phantom{0} dt \\ & = {t^2 \over 2} - 9t + c \\ \\ \text{When } t = 0 & \text{ and } v = -20, \phantom{000000} [\text{Initial velocity of -20 m/s}] \\ -20 & = {(0)^2 \over 2} - 9(0) + c \\ -20 & = 0 - 0 + c \\ -20 & = c \\ \\ \therefore v & = {t^2 \over 2} - 9t - 20 \\ v & = {1 \over 2}t^2 - 9t - 20 \end{align}

(ii)

\begin{align} v & = {1 \over 2}t^2 - 9t - 20 \\ \\ \text{When } & v = 0, \\ 0 & = {1 \over 2}t^2 - 9t - 20 \\ 0 & = t^2 - 18t - 40 \\ 0 & = (t - 20)(t + 2) \\ \\ t - 20 = 0 \phantom{000} & \text{or} \phantom{000} t + 2 = 0 \\ t = 20 \phantom{00} & \phantom{or000+2} t = -2 \text{ (Reject, } t \ge 0) \end{align}

(i)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int 8e^{-{1 \over 2}} \phantom{0} dt \\ & = 8e^{-{1 \over 2}} t + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \phantom{000000} [\text{Particle starts from } O] \\ 0 & = 8e^{-{1 \over 2}} (0) + c \\ 0 & = 0 + c \\ 0 & = c \\ \\ \therefore s & = (8e^{-{1 \over 2}} t) \text{ m} \end{align}

(ii)

\begin{align} s & = 8e^{-{1 \over 2}}t \\ \\ \text{When } & t = 5, \\ s & = 8e^{-{1 \over 2}} (5) \\ & = 40e^{-{1 \over 2}} \text{ m} \end{align}

(i)

\begin{align} s & = t(t - 6)^2 \\ & = t[ (t)^2 - 2(t)(6) + (6)^2 ] \\ & = t ( t^2 - 12t + 36 ) \\ & = t^3 - 12t^2 + 36t \\ \\ v & = {ds \over dt} \\ & = {d \over dt} (t^3 - 12t^2 + 36t) \\ & = (3)(t^2) - 12(2)(t) + 36(1) \\ & = 3t^2 - 24t + 36 \\ \\ \text{When } & t = 1, \\ v & = 3(1)^2 - 24(1) + 36 \\ & = 15 \text{ m/s} \end{align}

(ii)

\begin{align} v & = 3t^2 - 24t + 36 \\ \\ \text{When } & v = 0, \\ 0 & = 3t^2 - 24t + 36 \\ 0 & = t^2 - 8t + 12 \\ 0 & = (t - 2)(t - 6) \\ \\ t - 2 = 0 \phantom{000} & \text{or} \phantom{000} t - 6 = 0 \\ t = 2 \phantom{000} & \phantom{or000-6} t = 6 \end{align}

(iii)

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} (3t^2 - 24t + 36) \\ & = 3(2)(t) - 24(1) + 0 \\ & = 6t - 24 \\ \\ \text{When } & t = 3, \\ a & = 6(3) - 24 \\ & = -6 \text{ m/s}^2 \end{align}

(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} (12t - t^3) \\ & = 12(1) - (3)(t^2) \\ & = 12 - 3t^2 \\ \\ \text{When } & v = 0, \phantom{00000} [\text{At rest}] \\ 0 & = 12 - 3t^2 \\ 3t^2 & = 12 \\ t^2 & = {12 \over 3} \\ t^2 & = 4 \\ t & = \pm \sqrt{4} \\ & = 2 \phantom{0} \text{ or } \phantom{0} -2 \text{ (Reject, } t \ge 0) \\ \\ a & = {dv \over dt} \\ & = {d \over dt} (12 - 3t^2) \\ & = 0 - 3(2)(t) \\ & = -6t \\ \\ \text{When } & t = 2, \\ a & = -6(2) \\ & = -12 \text{ m/s}^2 \end{align}

(ii)

\begin{align} s & = 12t - t^3 \\ \\ \text{When } & s = 0, \\ 0 & = 12t - t^3 \\ 0 & = t(12 - t^2) \end{align} \begin{align} t & = 0 \phantom{000}& \text{ or} \phantom{000000} 12 - t^2 & = 0 \\ & & -t^2 & = -12 \\ & & t^2 & = 12 \\ & & t & = \pm \sqrt{12} \\ & & & = \sqrt{12} \phantom{0} \text{ or } \phantom{0} -\sqrt{12} \text{ (Reject, } t \ge 0 ) \end{align}
\begin{align} \text{Substitute } & t = \sqrt{12} \text{ into } v = 12 - 3t^2, \\ v & = 12 - 3(\sqrt{12})^2 \\ & = 12 - 3(12) \\ & = -24 \text{ m/s} \end{align}

(iii)

\begin{align} \text{From (i), } & \text{when } t = 0, s = 0 \\ \\ \text{From (i), } & v = 0 \text{ when } t = 2 \\ \\ s & = 12t - t^3 \\ \\ \text{When } & t = 2, \\ s & = 12(2) - (2)^3 \\ & = 16 \text{ m} \\ \\ \text{When } & t = 3, \\ s & = 12(3) - (3)^3 \\ & = 9 \text{ m} \end{align}

\begin{align} \text{Distance travelled} & = 16 + (16 - 9) \\ & = 23 \text{ m} \end{align}

(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} (18t^2 - t^4) \\ & = 18(2)(t) - (4)(t^3) \\ & = 36t - 4t^3 \\ \\ \text{When } & v = 0, \phantom{0000000} [\text{Particle at rest}] \\ 0 & = 36t - 4t^3 \\ 0 & = 9t - t^3 \\ 0 & = t(9 - t^2) \\ 0 & = t(3 - t)(3 + t) \end{align} \begin{align} t & = 0 \phantom{000} & \text{or} \phantom{000} 3 - t & = 0 \phantom{000} & \text{or} \phantom{000} 3 + t & = 0 \\ & & -t & = -3 & t & = -3 \text{ (Reject, } t \ge 0) \\ & & t & = 3 \end{align}
\begin{align} \text{Substitute } & t = 0 \text{ into } s = 18t^2 - t^4, \\ s & = 18(0)^2 - (0)^4 \\ & = 0 \phantom{00} \text{ (Reject, non-zero value of } s) \\ \\ \text{When } & t = 3 \text{ into } s = 18t^2 - t^4, \\ s & = 18(3)^2 - (3)^4 \\ & = 81 \end{align}

(ii) From part (i), we know that the particle comes is at rest when $t = 0$ and $t = 3$. Thus, the particle is travelling in one direction from $t = 0$ to $t = 3$.

For the the third second, we need to find the distance travelled from $t = 2$ to $t = 3$.

\begin{align} s & = 18t^2 - t^4, \\ \\ \text{When } & t = 2, \\ s & = 18(2)^2 - (2)^4 \\ & = 56 \text{ m} \\ \\ \text{When } & t = 3, \\ s & = 18(3)^2 - (3)^4 \\ & = 81 \text{ m} \\ \\ \therefore \text{Distance travelled} & = 81 - 56 \\ & = 25 \text{ m} \end{align}

(iii)

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} (36t - 4t^3) \\ & = 36(1) - 4(3)(t^2) \\ & = 36 - 12t^2 \\ \\ \text{When } & a = -12, \\ -12 & = 36 - 12t^2 \\ 0 & = 48 - 12t^2 \\ 0 & = 4 - t^2 \\ t^2 & = 4 \\ t & = \pm \sqrt{4} \\ t & = \pm 2 \\ \\ t = 2 \text{ or } & \phantom{.} t = - 2 \text{ (Reject, } t \ge 0) \\ \\ \text{Substitute } & t = 2 \text{ into } v = 36t - 4t^3, \\ v & = 36(2) - 4(2)^3 \\ & = 40 \text{ m/s} \end{align}

(i)

\begin{align} v & = {dh \over dt} \\ & = {d \over dt} (24t - 3t^2) \\ & = 24(1) - 3(2)(t) \\ & = 24 - 6t \\ \\ \text{When } & t = 3, \\ v & = 24 - 6(3) \\ & = 6 \text{ m/s} \end{align}

(ii) At the maximum height, the velocity of the stone is $0 \text{ m/s}$.

\begin{align} \text{Substitute } & v = 0 \text{ into } v = 24 - 6t, \\ 0 & = 24 - 6t \\ 6t & = 24 \\ t & = {24 \over 6} \\ & = 4 \\ \\ \text{Substitute } & t = 4 \text{ into } h = 24t - 3t^2, \\ h & = 24(4) - 3(4)^2 \\ & = 48 \text{ m} \end{align}

(iii) When the stone completes it's 'flight', it is back at it's starting position (aka the ground). Thus, displacement is $0 \text{ m}$.

\begin{align} \text{Substitute } & h = 0 \text{ into } h = 24 - 3t^2, \\ 0 & = 24t - 3t^2 \\ 0 & = 8t - t^2 \\ 0 & = t(8 - t) \\ \\ t = 0 \phantom{000} & \text{or} \phantom{000} 8 - t = 0 \\ & \phantom{or0008} -t = -8 \\ & \phantom{or0008()0} t = 8 \\ \\ t & = 0 \implies \text{Starting position} \\ t & = 8 \implies \text{Back at starting position} \\ \\ \therefore \text{Yes, particle } & \text{returns to } O \text{ after 8 seconds} \end{align}

(i)

\begin{align} v & = 3(e^{0.4t} + 2) \\ & = 3e^{0.4t} + 6 \\ \\ a & = {dv \over dt} \\ & = {d \over dt} (3e^{0.4t} + 6) \\ & = (3e^{0.4t}){d \over dt}(0.4t) + 0 \\ & = (3e^{0.4t})(0.4) \\ & = 1.2e^{0.4t} \\ \\ \text{When } & t = 0, \\ a & = 1.2e^{0.4(0)}\\ & = 1.2(1) \\ & = 1.2 \\ & = {6 \over 5} \text{ m/s}^2 \end{align}

(ii)

\begin{align} s & = \int v \phantom{.} dt \\ & = \int 3e^{0.4t} + 6 \phantom{.} dt \\ & = {3e^{0.4t} \over 0.4} + 6t + c \\ & = 7.5e^{0.4t} + 6t + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \phantom{000000} [\text{Initially at } O] \\ 0 & = 7.5e^{0.4(0)} + 6(0) + c \\ 0 & = 7.5(1) + 0 + c \\ 0 & = 7.5 + c \\ -7.5 & = c \\ \\ \therefore s & = 7.5e^{0.4t} + 6t - 7.5 \\ \\ \text{When } & t = 1, \\ s & = 7.5e^{0.4(1)} + 6(1) - 7.5 \\ & = 7.5e^{0.4} - 1.5 \\ & = {15 \over 2}e^{0.4} - {3 \over 2} \\ & = {1 \over 2} (15e^{0.4} - 3) \text{ m} \end{align}

(iii)

\begin{align} v & = 3e^{0.4t} + 6 \\ \\ a & = {dv \over dt} = 1.2e^{0.4t} \\ \\ \text{For } t \ge 0 , & \phantom{.} {dv \over dt} > 0 \text{ and } v \text{ is an increasing function} \\ \\ \implies \text{Parti} & \text{cle will never change direction} \\ \therefore \text{Particle } & \text{will not return to } O \end{align}

(i)

\begin{align} \text{When } & v = 0, \\ 0 & = 2t^2 - 3t - 2 \\ 0 & = (2t + 1)(t - 2) \end{align} \begin{align} t - 2 & = 0 &\text{or} \phantom{000} 2t + 1 & = 0 \\ t & = 2 & 2t & = -1\\ & & t & = -{1 \over 2} \text{ (Reject, } t \ge 0 ) \end{align}
\begin{align} s & = \int v \phantom{0} dt \\ & = \int (2t^2 - 3t - 2) \phantom{0} dt \\ & = {2t^3 \over 3} - {3t^2 \over 2} - 2t + c \\ \\ \text{When } & t = 0 \text{ and } s = 3, \\ 3 & = {2(0)^3 \over 3} - {3(0)^2 \over 2} - 2(0) + c \\ 3 & = 0 - 0 - 0 + c \\ 3 & = c \\ \\ \therefore s & = {2t^3 \over 3} - {3t^2 \over 2} - 2t + 3 \\ \\ \text{When } & t = 2, \\ s & = {2(2)^3 \over 3} - {3(2)^2 \over 2} - 2(2) + 3 \\ & = -1{2 \over 3} \text{ m} \end{align}

(ii)

\begin{align} \text{From (i), when } & t = 0, \phantom{.} s = 3 \text{ m} \\ \text{From (i), when } & t = 2, \phantom{.} s = -1{2 \over 3} \text{ m} \\ \\ s & = {2t^3 \over 3} - {3t^2 \over 2} - 2t + 3 \\ \\ \text{When } & t = 3, \\ s & = {2(3)^3 \over 3} - {3(3)^2 \over 2} - 2(3) + 3 \\ & = 1{1 \over 2} \text{ m} \end{align}

\begin{align} \therefore \text{Total distance } & = \left[ 3 - \left(-1{2 \over 3}\right) \right] + \left[ 1{1 \over 2} - \left(-1{2 \over 3} \right) \right] \\ & = 4{2 \over 3} + 3{1 \over 6} \\ & = 7{5 \over 6} \text{ m} \end{align}

(i)

\begin{align} a & = 2(3 - t) \\ & = 6 - 2t \\ \\ v & = \int (6 - 2t) \phantom{0} dt \\ & = 6t - {2t^2 \over 2} + c \\ & = 6t - t^2 + c \\ \\ \text{When } t = 0 & \text{ and } v = 7, \phantom{000000} [\text{Particle passes } O \text{ with } v = 7] \\ 7 & = 6(0) - (0)^2 + c \\ 7 & = 0 - 0 + c \\ 7 & = c \\ \\ \therefore v & = 6t - t^2 + 7 \\ \\ \text{When } & v = 7, \\ 7 & = 6t - t^2 + 7 \\ 0 & = 6t - t^2 \\ 0 & = t(6 - t) \end{align} \begin{align} t & = 0 \text{ (Repeated)} & \text{or} \phantom{000} 6 - t & = 0 \\ & & -t & = -6 \\ & & t & = 6 \end{align}

(ii)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int (6t - t^2 + 7) \phantom{0} dt \\ & = {6t^2 \over 2} - {t^3 \over 3} + 7t + c \\ & = 3t^2 - {1 \over 3}t^3 + 7t + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \\ 0 & = 3(0)^2 - {1 \over 3}(0)^3 + 7(0) + c \\ 0 & = 0 - 0 + 0 + c \\ 0 & = c \\ \\ \therefore s & = 3t^2 - {1 \over 3}t^3 + 7t \\ \\ \text{When } & t = 6, \\ s & = 3(6)^2 - {1 \over 3}(6)^3 + 7(6) \\ & = 78 \\ \\ \therefore \text{Particle is} & \text{ 78 m from point } O \end{align}

(i)

\begin{align} v & = \int a \phantom{0} dt \\ & = \int (8 - 4t) \phantom{0} dt \\ & = 8t - {4t^2 \over 2} + c \\ & = 8t - 2t^2 + c \\ \\ \text{When } t = 0 & \text{ and } v = 0, \phantom{000000} [\text{Particle initially at rest}] \\ 0 & = 8(0) - 2(0)^2 + c \\ 0 & = 0 - 0 + c \\ 0 & = c \\ \\ \therefore v & = 8t - 2t^2 \\ \\ \text{When } & v = 0, \phantom{000000} [\text{Particle instantaneously at rest}] \\ 0 & = 8t - 2t^2 \\ 0 & = 4t - t^2 \\ 0 & = t(4 - t) \end{align} \begin{align} t & = 0 \text{ (Repeated)} & \text{or} \phantom{000} 4 - t & = 0 \\ & & -t & = -4 \\ & & t & = 4 \end{align}

(ii)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int (8t - 2t^2) \phantom{0} dt \\ & = {8t^2 \over 2} - {2t^3 \over 3} + c \\ & = 4t^2 - {2 \over 3}t^3 + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \phantom{000000} [\text{Particle initially at } O] \\ 0 & = 4(0)^2 - {2 \over 3}(0)^3 + c \\ 0 & = 0 - 0 + c \\ 0 & = c \\ \\ \therefore s & = 4t^2 - {2 \over 3}t^3 \\ \\ \text{When } & t = 4, \phantom{000000} [\text{From (i), } v = 0 \text{ when } t = 4] \\ s & = 4(4)^2 - {2 \over 3}(4)^3 \\ & = 21{1 \over 3} \\ \\ \therefore OA & = 21{1 \over 3} \text{ m} \end{align}

(iii)

\begin{align} {dv \over dt} & = a = 8 - 4t \\ \\ \text{When } & a = 0, \\ 0 & = 8 - 4t \\ 4t & = 8 \\ t & = {8 \over 4} \\ & = 2 \\ \\ \text{Substitute } & t = 2 \text{ into } v = 8t - 2t^2, \\ v & = 8(2) - 2(2)^2 \\ & = 8 \text{ m/s} \\ \\ {d^2 v \over dt^2} & = {d \over dt} (8 - 4t) \\ & = - 4 < 0 \\ \\ \therefore \text{Max.} & \text{ speed is } 8 \text{ m/s} \end{align}

(i)

\begin{align} v & = \int a \phantom{0} dt \\ & = \int (13 - 6t) \phantom{0} dt \\ & = 13t - {6t^2 \over 2} + c \\ & = 13t - 3t^2 + c \\ \\ \text{When } t = 0 & \text{ and } v = 30, \phantom{000000} [\text{Initial velocity of 30 m/s}] \\ 30 & = 13(0) - 3(0)^2 + c \\ 30 & = 0 - 0 + c \\ 30 & = c \\ \\ \therefore v & = 13t - 3t^2 + 30 \\ \\ \text{When } & t = 3, \\ v & = 13(3) - 3(3)^2 + 30 \\ & = 42 \text{ m/s} \end{align}

(ii)

\begin{align} {ds \over dt} & = v = 13t - 3t^2 + 30 \\ \\ \text{When } & v = 0,\\ 0 & = 13t - 3t^2 + 30 \\ 0 & = -3t^2 + 13t + 30 \\ 0 & = 3t^2 - 13t - 30 \\ 0 & = (3t + 5)(t - 6) \end{align} \begin{align} 3t + 5 & = 0 & \text{or} \phantom{000} t - 6 & = 0 \\ 3t & = -5 & t & = 6 \\ t & = -{5 \over 3} \text{ (Reject, } t \ge 0 ) \end{align}
\begin{align} {d^2 s \over dt^2} & = a = 13 - 6t \\ \\ \text{When } & t = 6, \\ {d^2 s \over dt^2} & = 13 - 6(6) \\ & = -23 < 0 \\ \\ \therefore \text{Max. dis} & \text{tance reached when } t = 6 \end{align}

(iii)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int (13t - 3t^2 + 30) \phantom{0} dt \\ & = {13t^2 \over 2} - {3t^3 \over 3} + 30t + c \\ & = {13 \over 2} t^2 - t^3 + 30t + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \phantom{000000} [\text{Particle initially at } O ] \\ 0 & = {13 \over 2} (0)^2 - (0)^3 + 30(0) + c \\ 0 & = 0 - 0 + 0 + c \\ 0 & = c \\ \\ \therefore s & = {13 \over 2}t^2 - t^3 + 30t \\ \\ \text{When } & t = 6, \phantom{0000000000} [\text{from (ii)}] \\ s & = {13 \over 2}(6)^2 - (6)^3 + 30(6) \\ & = 198 \\ \\ \therefore \text{Max.} & \text{ distance} = 198 \text{ m} \end{align}

(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} (t^3 - 9t^2 + 15t + 40) \\ & = (3)(t^2) - 9(2)(t) + 15(1) + 0 \\ & = 3t^2 - 18t + 15 \\ \\ \text{When } & v = 0, \\ 0 & = 3t^2 - 18t + 15 \\ 0 & = t^2 - 6t + 5 \\ 0 & = (t - 1)(t - 5) \\ \\ t - 1 = 0 \phantom{000} & \text{or} \phantom{000} t - 5 = 0 \\ t = 1 \phantom{000} & \phantom{or000-5} t = 5 \\ \\ \text{Substitute } & t = 1 \text{ into } s = t^3 - 9t^2 + 15t + 40, \\ s & = (1)^3 - 9(1)^2 + 15(1) + 40 \\ & = 47 \text{ m} \\ \\ \text{Substitute } & t = 5 \text{ into } s = t^3 - 9t^2 + 15t + 40, \\ s & = (5)^3 - 9(5)^2 + 15(5) + 40 \\ & = 15 \text{ m} \end{align}

(ii) The magnitude of the acceleration of the particle is $9 \text{ m/s}^2$. So the particle can be accelerating or decelerating. Thus, $a = 9$ or $a = - 9$.

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} (3t^2 - 18t + 15) \\ & = 3(2)(t) - 18(1) + 0 \\ & = 6t - 18 \\ \\ \text{When } & a = 9, \\ 9 & = 6t - 18 \\ 27 & = 6t \\ \\ 6t & = 27 \\ t & = {27 \over 6} \\ & = 4.5 \\ \\ \\ \text{When } & a = -9, \\ -9 & = 6t - 18 \\ 9 & = 6t \\ \\ 6t & = 9 \\ t & = {9 \over 6} \\ & = 1.5 \end{align}

(iii)

\begin{align} \text{Average speed} & = {\text{Total distance} \over \text{Total time}} \\ \\ \text{Substitute } & t = 0 \text{ into } s = t^3 - 9t^2 + 15t + 40, \\ s & = (0)^3 - 9(0)^2 + 15(0) + 40 \\ & = 40 \text{ m} \\ \\ \text{From (i), } & \text{when } t = 1, s = 47 \text{ m} \\ \\ \text{Substitute } & t = 2 \text{ into } s = t^3 - 9t^2 + 15t + 40, \\ s & = (2)^3 - 9(2)^2 + 15(2) + 40 \\ & = 42 \text{ m} \end{align}

\begin{align} \therefore \text{Total distance } & = (47 - 40) + (47 - 42) \\ & = 12 \text{ m} \\ \\ \text{Average speed } & = {12 \over 2} \\ & = 6 \text{ m/s} \end{align}

(iv)

\begin{align} \text{From (iii), } & \text{when } t = 0, s = 42 \text{ m} \\ \\ \text{From (i), } & \text{when } t = 1, s = 47 \text{ m} \\ \\ \text{From (i), } & \text{when } t = 5, s = 15 \text{ m} \\ \\ \text{Substitute } & t = 6 \text{ into } s = t^3 - 9t^2 + 15t + 40, \\ s & = (6)^3 - 9(6)^2 + 15(6) + 40 \\ & = 22 \text{ m} \end{align}

(The particle changes direction twice)

\begin{align} \therefore \text{Total distance } & = (47 - 40) + (47 - 15) + (22 - 15) \\ & = 46 \text{ m} \end{align}

(i)

\begin{align} v & = \int a \phantom{0} dt \\ & = \int (6t - 8) \phantom{0} dt \\ & = {6t^2 \over 2} - 8t + c \\ & = 3t^2 - 8t + c \\ \\ \text{When } t = 0 & \text{ and } v = 4, \phantom{000000} [\text{Initial velocity is 4 m/s}] \\ 4 & = 3(0)^2 - 8(0) + c \\ 4 & = 0 - 0 + c \\ 4 & = c \\ \\ \therefore v & = 3t^2 - 8t + 4 \\ \\ \text{When } & v = 0, \\ 0 & = 3t^2 - 8t + 4 \\ 0 & = (3t - 2)(t - 2) \end{align} \begin{align} 3t - 2 & = 0 & \text{or} \phantom{000} t - 2 & = 0 \\ 3t & = 2 & t & = 2 \\ t & = {2 \over 3} \end{align}
Thus, the particle first comes to rest at $t = {2 \over 3}$
\begin{align} s & = \int v \phantom{0} dt \\ & = \int (3t^2 - 8t + 4) \phantom{0} dt \\ & = {3t^3 \over 3} - {8t^2 \over 2} + 4t + c \\ & = t^3 - 4t^2 + 4t + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \phantom{000000} [\text{Initially at } O] \\ 0 & = (0)^3 - 4(0)^2 + 4(0) + c \\ 0 & = 0 - 0 + 0 + c \\ 0 & = c \\ \\ \therefore s & = t^3 - 4t^2 + 4t \\ \\ \text{When } & t = {2 \over 3}, \\ s & = \left({2 \over 3}\right)^3 - 4\left(2 \over 3\right)^2 + 4\left(2 \over 3\right) \\ & = {32 \over 27} \text{ m} \end{align}

(ii)

\begin{align} s & = t^3 - 4t^2 + 4t \\ \\ \text{When } & s = 0, \\ 0 & = t^3 - 4t^2 + 4t \\ 0 & = t(t^2 - 4t + 4) \\ 0 & = t(t - 2)^2 \end{align} \begin{align} t & = 0 & \text{or} \phantom{000} t - 2 & = 0 \\ & & t & = 2 \end{align}
$$ \text{Time particle takes to return to } O = 2 \text{ seconds} $$

(iii)

\begin{align} {dv \over dt} & = a = 6t - 8 \\ \\ \text{When } & a = 0, \\ 0 & = 6t - 8 \\ 8 & = 6t \\ \\ 6t & = 8 \\ t & = {8 \over 6} \\ & = {4 \over 3} \\ \\ \text{Substitute } & t = {4 \over 3} \text{ into } v = 3t^2 - 8t + 4, \\ v & = 3\left(4 \over 3\right)^2 - 8\left(4 \over 3\right) + 4 \\ & = -{4 \over 3} \text{ m/s} \\ \\ {d^2 v \over dt^2} & = 6(1) \\ & = 6 > 0 \\ \\ \therefore \text{Minimum } & \text{velocity is } -{4 \over 3} \text{ m/s} \end{align}

(i)

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} (-3t^2 + 8t + 5) \\ & = -3(2)(t) + 8(1) + 0 \\ & = - 6t + 8 \\ & = (8 - 6t) \text{ m/s}^2 \end{align}

(ii)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int (-3t^2 + 8t + 5) \phantom{0} dt \\ & = -{3t^3 \over 3} + {8t^2 \over 2} + 5t + c \\ & = -t^3 + 4t^2 + 5t + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \phantom{000000} [\text{Particle starts from } O] \\ 0 & = -(0)^3 + 4(0)^2 + 5(0) + c \\ 0 & = -0 + 0 + 0 + c \\ 0 & = c \\ \\ \therefore s & = (-t^3 + 4t^2 + 5t) \text{ m} \end{align}

(iii) From Physics: Speed is a scalar quantity which has only magnitude while velocity is vector quantity which has magnitude and direction. Thus, if the speed of the particle is $2 \text{ m/s}$, the velocity of the particle can be either $2 \text{ m/s}$ or $-2 \text{ m/s}$.

\begin{align} \text{Substitute } & v = 2 \text{ into } v = -3t^2 + 8t + 5, \\ 2 & = -3t^2 + 8t + 5 \\ 0 & = -3t^2 + 8t + 3 \\ 0 & = 3t^2 - 8t - 3 \\ 0 & = (3t + 1)(t - 3) \end{align} \begin{align} 3t + 1 & = 0 & \text{or} \phantom{000} t - 3 & = 0 \\ 3t & = - 1 & t & = 3 \\ t & = -{1 \over 3} \text{ (Reject)} \end{align}
\begin{align} \text{Substitute } & v = -2 \text{ into } v = -3t^2 + 8t + 5, \\ -2 & = -3t^2 + 8t + 5 \\ 0 & = -3t^2 + 8t + 5 + 2 \\ 0 & = -3t^2 + 8t + 7 \\ 0 & = 3t^2 - 8t - 7 \end{align} \begin{align} t & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-8) \pm \sqrt{(-8)^2 - 4(3)(-7)} \over 2(3)} \\ & = {8 \pm \sqrt{148} \over 6} \\ & = {8 \pm \sqrt{4} \sqrt{37} \over 6} \\ & = {8 \pm 2\sqrt{37} \over 6} \\ & = {4 \pm \sqrt{37} \over 3} \\ & = {1 \over 3}(4 \pm \sqrt{37}) \\ & = {1 \over 3}(4 + \sqrt{37}) \phantom{0} \text{ or } \phantom{0} {1 \over 3}(4 - \sqrt{37}) \text{ (Reject, } t \ge 0 ) \end{align}

(iv)

\begin{align} \text{Substitute } & s = 20 \text{ into } s = -t^3 + 4t^2 + 5t, \\ 20 & = -t^3 + 4t^2 + 5t \\ 0 & = -t^3 + 4t^2 + 5t - 20 \\ 0 & = t^3 - 4t^2 - 5t + 20 \\ \\ \text{Let } f(t) & = t^3 - 4t^2 - 5t + 20 \\ \\ f(4) & = (4)^3 - 4(4)^2 - 5(4) + 20 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } t - 4 \text{ is a factor} \end{align}
$$ \require{enclose} \begin{array}{rll} t^2 - 5\phantom{0000000}\\ t - 4 \enclose{longdiv}{t^3 - 4t^2 - 5t + 20\phantom{0}}\kern-.2ex \\ -\underline{(t^3 - 4t^2){\phantom{00000000}}} \\ -5t + 20 \phantom{0} \\ -\underline{(-5t + 20){\phantom{}}} \\ 0\phantom{0} \end{array} $$
\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ t^3 - 4t^2 - 5t + 20 & = (t - 4)(t^2 - 5) + 0 \\ & = (t - 4)(t^2 - 5) \\ \\ 0 & = t^3 - 4t^2 - 5t + 20 \\ 0 & = (t - 4)(t^2 - 5) \end{align} \begin{align} t - 4 & = 0 & \text{or} \phantom{000} t^2 - 5 & = 0 \\ t & = 4 & t^2 & = 5 \\ & & t & = \pm\sqrt{5} \\ & & & = \sqrt{5} \phantom{0} \text{ or } \phantom{0} -\sqrt{5} \text{ (Reject, } t \ge 0 ) \end{align}

(i)

\begin{align} \text{When } & v = 0, \\ 0 & = 9 - t^2 \\ t^2 & = 9 \\ t & = \pm \sqrt{9} \\ & = \pm 3 \\ & = 3 \phantom{0} \text{ or } -3 \text{ (Reject, } t \ge 0) \end{align}

(ii)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int (9 - t^2) \phantom{0} dt \\ & = 9t - {t^3 \over 3} + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \phantom{000000} [\text{Particle starts from starting point}] \\ 0 & = 9(0) - {(0)^3 \over 3} + c \\ 0 & = 0 - 0 + c \\ 0 & = c \\ \\ \therefore s & = 9t - {t^3 \over 3} \\ \\ \text{When } & s = 0, \\ 0 & = 9t - {t^3 \over 3} \\ 0 & = 27t - t^3 \\ 0 & = t(27 - t^2) \end{align} \begin{align} t & = 0 & \text{or} \phantom{000} 27 - t^2 & = 0 \\ & & -t^2 & = -27 \\ & & t^2 & = 27 \\ & & t & = \pm\sqrt{27} \\ & & & = \pm \sqrt{9}\sqrt{3} \\ & & & = \pm 3\sqrt{3} \\ & & & = 3\sqrt{3} \phantom{0} \text{ or } -3\sqrt{3} \text{ (Reject, } t \ge 0) \end{align}
\begin{align} \text{Substitute } & t = 3\sqrt{3} \text{ into } v = 9 - t^2, \\ v & = 9 - (3\sqrt{3})^2 \\ & = -18 \\ \\ \therefore \text{Speed} & = 18 \text{ m/s} \end{align}

(iii)

\begin{align} \text{When } & s = 18, \\ 18 & = 9t - {t^3 \over 3} \\ 0 & = -{t^3 \over 3} + 9t - 18 \\ 0 & = -t^3 + 27t - 54 \\ 0 & = t^3 - 27t + 54 \\ \\ \text{Let } f(t) & = t^3 - 27t + 54 \\ \\ f(3) & = (3)^3 - 27(3) + 54 \\ & = 0 \\ \\ \text{By Factor } & \text{theorem, } t - 3 \text{ is a factor} \end{align} $$ \require{enclose} \begin{array}{rll} t^2 + 3t - 18 \phantom{00}\\ t - 3 \enclose{longdiv}{t^3 - 0t^2 - 27t + 54\phantom{0}}\kern-.2ex \\ -\underline{(t^3 - 3t^2){\phantom{000000000}}} \\ 3t^2 - 27t + 54 \phantom{0} \\ -\underline{(3t^2 - 9t){\phantom{00000}}} \\ -18t + 54\phantom{0} \\ -\underline{(-18t + 54)} \\ 0\phantom{0} \end{array} $$
\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ t^3 - 27t + 54 & = (t - 3)(t^2 + 3t - 18) + 0 \\ & = (t - 3)(t^2 + 3t - 18) \\ & = (t - 3)(t - 6)(t + 3) \\ \\ 0 & = t^3 - 27t + 54 \\ 0 & = (t - 3)(t - 6)(t + 3) \end{align} \begin{align} t - 3 & = 0 & \text{or} \phantom{000} t - 6 & = 0 & \text{or} \phantom{000} t + 3 & = 0 \\ t & = 3 & t & = 6 & t & = -3 \text{ (Reject, } t \ge 0) \end{align}

\begin{align} \text{Maximum velocity} & \implies {dv \over dt} = 0 \\ & \implies a = 0 \text{ m/s}^2 \end{align}