Answer: \(\mathbf{v}_A=\binom{\frac{5\sqrt3}{2}}{\frac{15}{2}}\), \(\mathbf{v}_B=\binom{5\sqrt3}{5}\), and the boats collide when \(t=200\) seconds after boat \(A\) starts.

(a) Bearings are measured clockwise from north. Since the \(x\)-direction is east and the \(y\)-direction is north, a speed \(V\) on a bearing \(\alpha\) has velocity vector

\(\binom{V\sin\alpha}{V\cos\alpha}.\)

For boat \(A\), \(V=5\sqrt3\) and \(\alpha=30^\circ\), so

\(\mathbf v_A=\binom{5\sqrt3\sin30^\circ}{5\sqrt3\cos30^\circ} =\binom{\frac{5\sqrt3}{2}}{\frac{15}{2}}.\)

For boat \(B\), \(V=10\) and \(\alpha=60^\circ\), so

\(\mathbf v_B=\binom{10\sin60^\circ}{10\cos60^\circ} =\binom{5\sqrt3}{5}.\)

(b) Let \(t\) be the time in seconds after boat \(A\) starts.

The position vector of boat \(A\) is

\(\mathbf r_A=t\binom{\frac{5\sqrt3}{2}}{\frac{15}{2}}.\)

Boat \(B\) starts after \(100\) seconds from \(\binom0{1000}\). Therefore, for \(t\geqslant100\), its position vector is

\(\mathbf r_B=\binom0{1000}+(t-100)\binom{5\sqrt3}{5}.\)

For a collision, the \(x\)-coordinates and \(y\)-coordinates must be equal at the same time.

Equating the \(x\)-coordinates:

\(\dfrac{5\sqrt3}{2}t=5\sqrt3(t-100).\)

Divide by \(5\sqrt3\):

\(\dfrac t2=t-100.\)

So

\(t=200.\)

Now equate the \(y\)-coordinates:

\(\dfrac{15}{2}t=1000+5(t-100).\)

Simplifying the right-hand side,

\(\dfrac{15}{2}t=5t+500.\)

Multiplying by \(2\),

\(15t=10t+1000,\)

so

\(t=200.\)

Both coordinates give the same time, and \(200\geqslant100\), so boat \(B\) has already started. Therefore the two boats collide \(200\) seconds after boat \(A\) starts.