Answer: \(\mathbf v_A=5\mathbf i+5\sqrt3\mathbf j\); \(\mathbf r_A=t(5\mathbf i+5\sqrt3\mathbf j)\); \(\mathbf r_B=(2\sqrt3+\frac53t)\mathbf i+9\mathbf j\); they collide at \(t=\frac{3\sqrt3}{5}\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

Particle \(A\) has speed \(10\text{ m s}^{-1}\) at \(60^\circ\) above the \(x\)-axis.

Its velocity components are

\(10\cos60^\circ=5\) and \(10\sin60^\circ=5\sqrt3.\)

So \(\mathbf v_A=5\mathbf i+5\sqrt3\mathbf j\).

Since \(A\) starts from the origin, its position vector at time \(t\) is

\(\mathbf r_A=5t\mathbf i+5\sqrt3t\mathbf j.\)

Particle \(B\) starts at \((2\sqrt3,9)\) and moves parallel to the positive \(x\)-axis with speed \(\frac53\). Hence

\(\mathbf r_B=\left(2\sqrt3+\frac53t\right)\mathbf i+9\mathbf j.\)

For a collision, the \(y\)-coordinates must be equal:

\(5\sqrt3t=9\), so \(t=\frac{9}{5\sqrt3}=\frac{3\sqrt3}{5}.\)

At this time, the \(x\)-coordinate of \(A\) is

\(5t=3\sqrt3.\)

The \(x\)-coordinate of \(B\) is

\(2\sqrt3+\frac53\cdot\frac{3\sqrt3}{5}=2\sqrt3+\sqrt3=3\sqrt3.\)

Both coordinates match at the same time, so the particles collide.

This gives the required answer: \(\mathbf v_A=5\mathbf i+5\sqrt3\mathbf j\); \(\mathbf r_A=t(5\mathbf i+5\sqrt3\mathbf j)\); \(\mathbf r_B=(2\sqrt3+\frac53t)\mathbf i+9\mathbf j\); they collide at \(t=\frac{3\sqrt3}{5}\).