(a) Forces \(F_{1} = \begin{pmatrix} -5 & 4 \end{pmatrix} N; F_{2} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} N; F_{3} = \begin{pmatrix} 2 & -1 \end{pmatrix} N\...

(a)

(i) Resultant force

Writing each force as a column vector,

\[\begin{aligned}\mathbf R&=\begin{pmatrix}-5\\4\end{pmatrix}+\begin{pmatrix}2\\5\end{pmatrix}+\begin{pmatrix}2\\-1\end{pmatrix}+\begin{pmatrix}3\\-5\end{pmatrix}\\&=\begin{pmatrix}-5+2+2+3\\4+5-1-5\end{pmatrix}\\&=\begin{pmatrix}2\\3\end{pmatrix}\text{ N}.\end{aligned}\]

Hence, \(\boxed{\mathbf R=2\mathbf i+3\mathbf j\text{ N}}\).

(ii) Fifth force for equilibrium

For equilibrium, the fifth force \(\mathbf F_5\) must be the negative of the resultant:

\[\mathbf F_5=-\mathbf R=-\begin{pmatrix}2\\3\end{pmatrix}=\boxed{\begin{pmatrix}-2\\-3\end{pmatrix}\text{ N}}.\]

(b)

During the first 4 s,

\[v=20+(2.5)(4)=30\text{ m s}^{-1}.\]

Let the magnitude of the retardation be \(r\text{ m s}^{-2}\). Since

\[2.5:r=3:4,\qquad r=\frac{4}{3}(2.5)=\frac{10}{3}\text{ m s}^{-2}.\]

(i) Velocity-time graph

The graph consists of straight-line sections joining \((0,20)\), \((4,30)\), \((12,30)\), and \((21,0)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).

(ii) Time taken to come to rest

For the retardation stage, \(u=30\text{ m s}^{-1}\), \(v=0\), and acceleration \(=-\frac{10}{3}\text{ m s}^{-2}\).

\[0=30-\frac{10}{3}t\]

\[\frac{10}{3}t=30\]

\[\boxed{t=9\text{ s}}.\]

(iii) Total distance travelled

The total distance is the area under the velocity-time graph:

\[\begin{aligned}A_1&=\frac12(20+30)\times4=100\text{ m},\\A_2&=30\times8=240\text{ m},\\A_3&=\frac12\times30\times9=135\text{ m}.\end{aligned}\]

\[\text{Total distance}=100+240+135=\boxed{475\text{ m}}.\]