Answer: (a) \(\begin{pmatrix}-40\\42\end{pmatrix}\). (b) \(\begin{pmatrix}5-40t\\-3+42t\end{pmatrix}\). (c) \(\begin{pmatrix}5t-1\\2t+1\end{pmatrix}\). (d) \(\sqrt{29t^2-6t+2}\). (e) \(t=0.49\).

Answer: (a) \(\begin{pmatrix}-40\\42\end{pmatrix}\). (b) \(\begin{pmatrix}5-40t\\-3+42t\end{pmatrix}\). (c) \(\begin{pmatrix}5t-1\\2t+1\end{pmatrix}\). (d) \(\sqrt{29t^2-6t+2}\). (e) \(t=0.49\).

(a) The direction vector \(\begin{pmatrix}-20\\21\end{pmatrix}\) has magnitude

\(\sqrt{(-20)^2+21^2}=\sqrt{841}=29\).

A speed of 58 is twice this magnitude, so the velocity vector is

\(\displaystyle 2\begin{pmatrix}-20\\21\end{pmatrix}=\begin{pmatrix}-40\\42\end{pmatrix}\).

(b) Starting from \(\begin{pmatrix}5\\-3\end{pmatrix}\), the position vector of \(A\) at time \(t\) is

(c) The displacement vector \(\overrightarrow{AB}\) is the position vector of \(B\) minus the position vector of \(A\):

(d) Therefore

\(\displaystyle AB=\sqrt{(5t-1)^2+(2t+1)^2}\).

Expanding,

\(AB=\sqrt{25t^2-10t+1+4t^2+4t+1}\),

so

\(AB=\sqrt{29t^2-6t+2}\).

(e) If \(AB=\sqrt6\), then

\(29t^2-6t+2=6\).

So

\(29t^2-6t-4=0\).

Solving,

\(\displaystyle t=\frac{6\pm\sqrt{36+464}}{58}=\frac{6\pm\sqrt{500}}{58}\).

The positive value is

\(t\approx0.49\).