(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} (27t - t^3) \\ & = 27(1) - (3)(t^2) \\ & = 27 - 3t^2 \\ \\ \text{When } & v = 0, \phantom{000000} [\text{Instantaneously at rest}] \\ 0 & = 27 - 3t^2 \\ 3t^2 & = 27 \\ t^2 & = 9 \\ t & = \pm \sqrt{9} \\ & = \pm 3 \\ & = 3 \phantom{0} \text{ or } - 3 \text{ (Reject, } t \ge 0 ) \\ \\ a & = {dv \over dt} \\ & = {d \over dt} (27 - 3t^2) \\ & = 0 - 3(2)(t) \\ & = -6t \\ \\ \text{When } & t = 3, \\ a & = -6(3) \\ & = -18 \text{ cm/s}^2 \end{align}

(ii) When the particle is next at O, the displacement of the particle is 0 m

\begin{align} \text{Substitute } & s = 0 \text{ into } s = 27t - t^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 = 27 - 3t^2, \\ v & = 27 - 3(3\sqrt{3})^2 \\ & = -54 \text{ cm/s} \end{align}

(iii) To find the distance travelled in the fourth second, we need to find the displacement of the particle in the third second and in the fourth second

\begin{align} \text{Substitute } & t = 3 \text{ into } s = 27t - t^3, \\ s & = 27(3) - (3)^3 \\ & = 54 \text{ cm} \\ \\ \text{Substitute } & t = 4 \text{ into } s = 27t - t^3, \\ s & = 27(4) - (4)^3 \\ & = 44 \text{ cm} \\ \\ \therefore \text{Distance travelled} & = 54 - 44 \\ & = 10 \text{ cm} \end{align}

(iv)

\begin{align} \text{Substitute } & t = 0 \text{ into } s = 27t - t^3, \\ s & = 27(0) - (0)^3 \\ & = 0 \text{ cm} \\ \\ \text{From (iii), } \text{when } & t = 3, s = 54 \text{ cm} \\ \\ \text{From (iii), } \text{when } & t = 4, s = 44 \text{ cm} \end{align}

\begin{align} \therefore \text{Distance travelled } & = 54 + (54 - 44) \\ & = 64 \text{ cm} \end{align}

(i) Speed is a scalar quantity that only has magnitude but no direction. Since the speed is 8 m/s, the velocity can be 8 m/s or -8 m/s.

\begin{align} \text{When } & v = \pm 8, \\ \pm 8 & = 12 \sin {t \over 2} \\ \pm {8 \over 12} & = \sin {t \over 2} \\ \pm {2 \over 3} & = \sin {t \over 2} \phantom{000000000} [\text{All 4 quadrants}] \\ \\ \text{Basic angle } & = \sin^{-1} \left(2 \over 3\right) \phantom{000000} [\text{Radian mode!}] \\ & = 0.72972 \end{align}

\begin{align} {t \over 2} & = 0.72972, \pi - 0.72972, \pi + 0.72972, 2\pi - 0.72972 \\ & = 0.72972, 2.4118, 3.8713, 5.5534 \\ \\ t & = 1.4594, 4.8237, 7.7426, 11.1068 \\ & \approx 1.46, 4.82, 7.74, 11.1 \\ \\ a & = {dv \over dt} \\ & = {d \over dt} \left(12 \sin {t \over 2} \right) \\ & = 12 \left(\cos {t \over 2} \right) {d \over dt} \left(t \over 2\right) \\ & = 12 \left(\cos {t \over 2} \right) \left(1 \over 2\right) \\ & = 6\cos {t \over 2} \\ \\ \text{When } & t = 1.4584, \\ a & = 6\cos \left({1.4594 \over 2}\right) \\ & = 4.4722 \\ & \approx 4.47 \text{ m/s}^2 \end{align}

(ii)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int \left(12 \sin {t \over 2} \right) \phantom{0} dt \\ & = {-12 \cos {t \over 2} \over {1 \over 2}} + c \\ & = - 24 \cos {t \over 2} + c \\ \\ \text{When } & t = 0 \text{ and } s = 0, \phantom{000000} [\text{Particle starts at } O] \\ 0 & = - 24 \cos \left({0 \over 2}\right) + c \\ 0 & = - 24(1) + c \\ 0 & = - 24 + c \\ 24 & = c \\ \\ \therefore s & = -24 \cos {t \over 2} + 24 \\ & = 24 - 24 \cos {t \over 2} \end{align}

(i)

\begin{align} u & = 3t &&& v & = t^2 + 9 \\ {du \over dt} & = 3 &&& {dv \over dt} & = 2t \end{align}
\begin{align} v & = {ds \over dt} \\ & = { (t^2 + 9)(3) - (3t)(2t) \over (t^2 + 9)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { 3(t^2 + 9) - 6t^2 \over (t^2 + 9)^2 } \\ & = { 3t^2 + 27 - 6t^2 \over (t^2 + 9)^2 } \\ & = { 27 - 3t^2 \over (t^2 + 9)^2 } \\ \\ \text{When } & t = 1, \\ v & = {27 - 3(1)^2 \over [(1)^2 + 9]^2} \\ & = {6 \over 25} \text{ m/s} \end{align}

(ii)

\begin{align} \text{Substitute } & t = 3 \text{ into } v = {27 - 3t^2 \over (t^2 + 9)^2}, \\ v & = {27 - 3(3)^2 \over [(3)^2 + 9]^2} \\ & = {0 \over 324} \\ & = 0 \text{ m/s} \\ \\ \text{Since } v = 0, \text{ the particle} & \text{ comes instantaneously to rest when } t = 3 \end{align}

(iii)

\begin{align} v & = {27 - 3t^2 \over (t^2 + 9)^2} \\ \\ \text{For } & t \ge 0, \phantom{.} (t^2 + 9)^2 > 0 \\ \\ \therefore 27 - 3t^2 & > 0 \\ 3t^2 - 27 & < 0 \\ t^2 - 9 & < 0 \\ (t - 3)(t + 3) < 0 \end{align}

\begin{align} -3 < \phantom{.} & t < 3 \\ \\ \text{Since } t \ge 0, & \phantom{.} 0 \le t < 3 \end{align}

(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} [12 \sin (2t) - 6] \\ & = 12(\cos 2t){d \over dt}(2t) - 0 \\ & = 12(\cos 2t)(2) - 0 \\ & = 24 \cos 2t \\ \\ a & = {dv \over dt} \\ & = {d \over dt} (24 \cos 2t) \\ & = 24(-\sin 2t) {d \over dt}(2t) \\ & = 24(-\sin 2t)(2) \\ & = -48 \sin 2t \end{align}

(ii)

\begin{align} \text{Substitute } & s = 0 \text{ into } s = 12\sin(2t)-6, \\ 0 & = 12 \sin (2t) - 6 \\ 6 & = 12 \sin (2t) \\ {6 \over 12} & = \sin (2t) \\ {1 \over 2} & = \sin (2t) \phantom{00000} [\text{1st & 2nd quadrants since } \sin (2t) > 0] \\ \\ \text{Basic angle } & = \sin^{-1} \left(1 \over 2\right) \\ & = 30^\circ \\ & = {\pi \over 6} \end{align}

\begin{align} 2t & = {\pi \over 6}, \pi - {\pi \over 6} \\ & = {\pi \over 6}, {5\pi \over 6} \\ \\ t & = {\pi \over 12}, {5\pi \over 12} \\ \\ \therefore \text{Particle first } & \text{reaches the fixed point at } {\pi \over 12} \text{ s} \end{align}

(iii) When the particle is at maximum distance from the fixed point, the particle is changing direction and is instantaneously at rest.

\begin{align} \text{Substitute } & v = 0 \text{ into } v = 24 \cos 2t, \\ 0 & = 24\cos 2t \\ 0 & = \cos 2t \end{align}

\begin{align} 2t & = {\pi \over 2}, {3\pi \over 2} \\ \\ t & = {\pi \over 4}, {3\pi \over 4} \\ \\ \text{Substitute } & t = {\pi \over 4} \text{ into } s = 12\sin(2t)-6, \\ s & = 12 \sin \left[ 2\left(\pi \over 4\right)\right] - 6 \\ & = 6 \text{ m} \\ \\ \text{Substitute } & t = {3\pi \over 4} \text{ into } s = 12\sin(2t)-6, \\ s & = 12 \sin \left[ 2\left(3\pi \over 4\right)\right] - 6 \\ & = -18 \text{ m} \\ \\ \text{Maximum dist} & \text{ance from the fixed point} = 18 \text{ m} \end{align}

(iv)

\begin{align} \text{Substitute } & s = 3 \text{ into } s = 12\sin(2t)-6, \\ 3 & = 12 \sin (2t) - 6 \\ 3 + 6 & = 12 \sin (2t) \\ 9 & = 12 \sin (2t) \\ {9 \over 12} & = \sin (2t) \\ {3 \over 4} & = \sin (2t)\\ \\ \sin^2 A + \cos^2 A & = 1 \\ \\ \sin^2 2t + \cos^2 2t & = 1 \\ \left(3 \over 4\right)^2 + \cos^2 (2t) & = 1 \\ {9 \over 16} + \cos^2 (2t) & = 1 \\ \cos^2 (2t) & = 1 - {9 \over 16} \\ \cos^2 (2t) & = {7 \over 16} \\ \cos (2t) & = \pm \sqrt{7 \over 16} \\ & = \pm {\sqrt{7} \over \sqrt{16}} \\ & = \pm {\sqrt{7} \over 4} \\ \\ v & = 24\cos 2t \\ & = 24 \left(\pm {\sqrt{7} \over 4} \right) \\ & = \pm 6\sqrt{7} \text{ m/s} \\ \\ \therefore \text{Speed} & = 6\sqrt{7} \text{ m/s} \end{align}

(i)

\begin{align} v & = \int a \phantom{0} dt \\ & = \int (2 - t) \phantom{0} dt \\ & = 2t - {t^2 \over 2} + c \\ \\ \text{When } & t = 0 \text{ and } v = 6, \phantom{000000} [\text{Initial velocity 6 m/s}] \\ 6 & = 2(0) - {(0)^2 \over 2} + c \\ 6 & = 0 - 0 + c \\ 6 & = c \\ \\ \therefore v & = 2t - {t^2 \over 2} + 6 \\ \\ \text{When } & t = 5, \\ v & = 2(5) - {(5)^2 \over 2} + 6 \\ & = 3.5 \text{ m/s} \end{align}

(ii)

(i)

\begin{align} a & = 2(3 - e^{-t}) \\ & = 6 - 2e^{-t} \\ \\ v & = \int a \phantom{0} dt \\ & = \int (6 - 2e^{-t}) \phantom{0} dt \\ & = 6t - {2e^{-t} \over -1} + c \\ & = 6t + 2e^{-t} + c \\ \\ \text{When } & t = 0 \text{ and } v = 0, \phantom{000000} [\text{Start from rest}] \\ 0 & = 6(0) + 2e^{-(0)} + c \\ 0 & = 0 + 2(1) + c \\ 0 & = 2 + c \\ -2 & = c \\ \\ \therefore v & = 6t + 2e^{-t} - 2 \\ \\ \text{When } & t = 2, \\ v & = 6(2) + 2e^{-(2)} - 2 \\ & = 10.2706 \\ & \approx 10.3 \text{ m/s} \end{align}

(ii)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int (6t + 2e^{-t} - 2) \phantom{0} dt \\ & = {6t^2 \over 2} + {2e^{-t} \over -1} - 2t + c \\ & = 3t^2 - 2e^{-t} - 2t + c \\ \\ \text{When } & t = 0 \text{ and } s = 0, \phantom{000000} [\text{Initially passes } O] \\ 0 & = 3(0)^2 - 2e^{-(0)} - 2(0) + c \\ 0 & = 0 - 2(1) - 0 + c \\ 0 & = - 2 + c \\ 2 & = c \\ \\ \therefore s & = 3t^2 - 2e^{-t} - 2t + 2 \\ \\ \text{When } & t = 1, \\ s & = 3(1)^2 - 2e^{-(1)} - 2(1) + 2 \\ & = 2.2642 \\ & \approx 2.26 \text{ m} \end{align}

(i)

\begin{align} \text{When } & t = 0 \text{ and } v = 12, \phantom{000000} [\text{Initial velocity 12 m/s}] \\ 12 & = {144 \over [2(0) + 3]^2} - 4k \\ 12 & = 16 - 4k \\ -4 & = -4k \\ {-4 \over -4} & = k \\ 1 & = k \end{align}

(ii)

\begin{align} \text{Since } k = 1, v & = {144 \over (2t + 3)^2} - 4(1) \\ & = {144 \over (2t + 3)^2} - 4 \\ \\ \text{When } & v = 0, \phantom{000000} [\text{Instantaneously at rest}] \\ 0 & = {144 \over (2t + 3)^2} - 4 \\ 4 & = {144 \over (2t + 3)^2} \\ 4(2t + 3)^2 & = 144 \\ (2t + 3)^2 & = {144 \over 4} \\ (2t + 3)^2 & = 36 \\ 2t + 3 & = \pm \sqrt{36} \\ 2t + 3 & = \pm 6 \end{align} \begin{align} 2t + 3 & = 6 & \text{or} \phantom{000} 2t + 3 & = -6 \\ 2t & = 3 & 2t & = - 9 \\ t & = {3 \over 2} & t & = -{9 \over 2} \text{ (Reject, } t \ge 0) \end{align}

(iii)

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} \left[ {144 \over (2t + 3)^2} - 4 \right] \\ & = {d \over dt} [ 144(2t + 3)^{-2} - 4 ] \\ & = \underbrace{ 144(-2)(2t + 3)^{-3} . (2)}_\text{Chain rule} - 0 \\ & = -576 (2t + 3)^{-3} \\ & = - {576 \over (2t + 3)^3} \\ \\ \text{When } & t = {1 \over 2}, \\ a & = -{576 \over [2\left(1 \over 2\right) + 3]^3} \\ & = -9 \text{ m/s}^2 \\ \\ \\ s & = \int v \phantom{0} dt \\ & = \int \left[ {144 \over (2t + 3)^2} - 4 \right] \phantom{0} dt \\ & = \int [ 144(2t + 3)^{-2} - 4 ] \phantom{0} dt \\ & = {144(2t + 3)^{-1} \over (-1)(2)} - 4t + c \\ & = -72(2t + 3)^{-1} - 4t + c \\ & = -{72 \over 2t + 3} - 4t + c \\ \\ \text{When } & t = 0 \text{ and } s = 0, \phantom{000000} [\text{Initially at } O] \\ 0 & = -{72 \over 2(0) + 3} - 4(0) + c \\ 0 & = -24 - 0 + c \\ 24 & = c \\ \\ \therefore s & = -{72 \over 2t + 3} - 4t + 24 \\ \\ \text{When } & t = {1 \over 2}, \\ s & = -{72 \over 2\left(1 \over 2\right) + 3} - 4\left(1 \over 2\right) + 24 \\ & = 4 \text{ m} \end{align}

(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} \left(4 - 4e^{-t} - {2 \over 5}t \right) \\ & = 0 - 4e^{-t}. (-1) - {2 \over 5} \\ & = 4e^{-t} - {2 \over 5} \\ \\ \text{When } & t = 0, \\ v & = 4e^{-(0)} - {2 \over 5} \\ & = 4(1) - {2 \over 5} \\ & = 3.6 \text{ m/s} \end{align}

(ii)

\begin{align} v & = 4e^{-t} - {2 \over 5} \\ \\ \text{When } & v = 0, \phantom{0000000000} [\text{Instantaneously at rest}] \\ 0 & = 4e^{-t} - {2 \over 5} \\ {2 \over 5} & = 4e^{-t} \\ {1 \over 10} & = e^{-t} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln {1 \over 10} & = \ln e^{-t} \\ \ln {1 \over 10} & = -t (\ln e) \phantom{0000000} [\text{Power law (logarithms)}] \\ \ln {1 \over 10} & = -t (1) \\ \ln {1 \over 10} & = -t \\ \\ t & = -\ln {1 \over 10} \\ & = 2.3025 \\ & \approx 2.30 \end{align}

(iii)

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} \left(4e^{-t} - {2 \over 5}\right) \\ & = 4e^{-t} {d \over dt}(-t) - 0 \\ & = 4e^{-t} (-1) \\ & = -4e^{-t} \\ \\ \text{When } & t = -\ln {1 \over 10}, \\ a & = -4e^{-(-\ln {1 \over 10})} \\ & = -4e^{\ln {1 \over 10}} \\ & = -4\left(1 \over 10\right) \\ & = -0.4 \text{ m/s}^2 \end{align}

(i)

\begin{align} v & = {ds \over dt} \\ & = {d \over dt} (21t^2 - 2t^3) \\ & = 21(2)(t) - 2(3)(t^2) \\ & = 42t - 6t^2 \\ \\ \text{When } & t = 1, \\ v & = 42(1) - 6(1)^2 \\ & = 36 \text{ m/s} \\ \\ \\ a & = {dv \over dt} \\ & = {d \over dt} (42t - 6t^2) \\ & = 42(1) - 6(2)(t) \\ & = 42 - 12t \\ \\ \text{When } & t = 1, \\ a & = 42 - 12(1) \\ & = 30 \text{ m/s}^2 \end{align}

(ii) When the particle changes direction, it is instantaneously at rest (i.e. v = 0) for an instant.

\begin{align} \text{Substitute } & v = 0 \text{ into } v = 42t - 6t^2, \\ 0 & = 42t - 6t^2 \\ 0 & = 3t(14 - 2t) \end{align} \begin{align} 3t & = 0 &\text{or} \phantom{000} 14 - 2t & = 0 \\ t & = 0 & -2t & = -14 \\ & & 2t & = 14 \\ & & t & = 7 \end{align}
$$ \therefore \text{Particle change its direction of motion when } t = 7 \text{ s}$$

(iii)

\begin{align} \text{Substitute } & t = 0 \text{ into } s = 21t^2 - 2t^3, \\ s & = 21(0)^2 - 2(0)^3 \\ & = 0 \text { m} \\ \\ \text{Substitute } & t = 7 \text{ into } s = 21t^2 - 2t^3, \phantom{000000} [\text{From ii, particle changes direction at } t = 7] \\ s & = 21(7)^2 - 2(7)^3 \\ & = 343 \text { m} \\ \\ \text{Substitute } & t = 10 \text{ into } s = 21t^2 - 2t^3, \\ s & = 21(10)^2 - 2(10)^3 \\ & = 100 \text { m} \end{align}

\begin{align} \therefore \text{Total distance } & = 343 + (343 - 100) \\ & = 586 \text{ m} \end{align}

(i)

\begin{align} s & = \int v \phantom{0} dt \\ & = \int (t^2 - 8t + 7) \phantom{0} dt \\ & = {t^3 \over 3} - {8t^2 \over 2} + 7t + c \\ & = {1 \over 3}t^3 - 4t^2 + 7t + c \\ \\ \text{When } & t = 0 \text{ and } s = 0, \phantom{000000} [\text{Passes } O \text{ initially}] \\ 0 & = {1 \over 3}(0)^3 - 4(0)^2 + 7(0) + c \\ 0 & = 0 - 0 + 0 + c \\ 0 & = c \\ \\ \therefore s & = {1 \over 3}t^3 - 4t^2 + 7t \end{align}

(ii) At points A and B, the particle is instantaneously at rest, which means it's velocity is 0 m/s.

\begin{align} \text{Substitute } & v = 0 \text{ into } v = t^2 - 8t + 7, \\ 0 & = t^2 - 8t + 7 \\ 0 & = (t - 1)(t - 7) \end{align} \begin{align} t - 1 & = 0 & \text{or} \phantom{000} t - 7 & = 0 \\ t & = 1 & t & = 7 \end{align}
\begin{align} \text{Substitute } & t = 1 \text{ into } s = {1 \over 3}t^3 - 4t^2 + 7t, \\ s & = {1 \over 3}(1)^3 - 4(1)^2 + 7(1) \\ & = 3{1 \over 3} \text{ m} \\ \\ \therefore \text{Point } A & \text{ is } 3{1 \over 3} \text{ m from } O \\ \\ \\ \text{Substitute } & t = 7 \text{ into } s = {1 \over 3}t^3 - 4t^2 + 7t, \\ s & = {1 \over 3}(7)^3 - 4(7)^2 + 7(7) \\ & = -32{2 \over 3} \text{ m} \\ \\ \therefore \text{Point } B & \text{ is } 32{2 \over 3} \text{ m from } O \phantom{.} (\text{in the opposite direction}) \\ \\ \therefore \text{Distance between }A \text{ and } B & = 3{1 \over 3} + 32{2 \over 3} \\ & = 36 \text{ m} \end{align}

(iii)

\begin{align} \text{When } & t = 0, s = 0 \text{ m} \\ \\ \text{From ii, when } & t = 1, s = 3{1 \over 3} \text{ m} \\ \\ \text{From ii, when } & t = 7, s = -32{2 \over 3} \text{ m} \\ \\ \text{Substitute } & t = 9 \text{ into } s = {1 \over 3}t^3 - 4t^2 + 7t, \\ s & = {1 \over 3}(9)^3 - 4(9)^2 + 7(9) \\ & = -18 \text{ m} \end{align}

(Thus, particle changes direction twice!)

\begin{align} \text{Total distance } & = 3{1 \over 3} + \left( 3{1 \over 3} + 32{2 \over 3} \right) + \left( 32{2 \over 3} - 18 \right) \\ & = 54 \text{ m} \\ \\ \\ a & = {dv \over dt} \\ & = {d \over dt} (t^2 - 8t + 7) \\ & = (2)(t) - 8(1) + 0 \\ & = 2t - 8 \\ \\ \text{When } & a = 0, \\ 0 & = 2t - 8 \\ 8 & = 2t \\ \\ t & = {8 \over 2} \\ & = 4 \\ \\ \text{Substitute } & t = 4 \text{ into } s = {1 \over 3}t^3 - 4t^2 + 7t, \\ s & = {1 \over 3}(4)^3 - 4(4)^2 + 7(4) \\ & = -14{2 \over 3} \text{ m} \\ \\ \text{From ii, } \text{when } & t = 7, s = -32{2 \over 3} \text{ m} \phantom{000000} [\text{Point } B] \\ \\ \therefore \text{Point } & C \text{ is closer to } O \end{align}

(i) Use the expression for the first four seconds

\begin{align} v & = \int (18 - 3t) \phantom{0} dt \\ & = 18t - {3t^2 \over 2} + c \\ \\ \text{When } t = 4 & \text{ and } v = 50, \\ 50 & = 18(4) - {3(4)^2 \over 2} + c \\ 50 & = 72 - 24 + c \\ 2 & = c \\ \\ \therefore v & = 18t - {3t^2 \over 2} + 2 \phantom{.} \text{ (For } 0 \le t < 4 ) \\ \\ \text{When } & t = 0, \\ v & = 18(0) - {3(0)^2 \over 2} + 2 \\ & = 0 - 0 + 2 \\ & = 2 \end{align}

(ii) Use the other expression!

\begin{align} v & = \int a \phantom{0} dt \\ & = \int (t + 2) \phantom{0} dt \\ & = {t^2 \over 2} + 2t + c \\ \\ \text{When } & t = 4 \text{ and } v = 50, \\ 50 & = {(4)^2 \over 2} + 2(4) + c \\ 50 & = 8 + 8 + c \\ 50 - 8 - 8 & = c \\ 34 & = c \\ \\ \therefore v & = {t^2 \over 2} + 2t + 34 \text{ (For } t > 4) \\ \\ \text{When } & t = 5, \\ v & = {(5)^2 \over 2} + 2(5) + 34 \\ & = 56.5 \text{ m/s} \end{align}

(iii)