Solution:

Let B be at \((0,0)\) and A at \((a,0)\) at \(t=0\). Boy B's position at time \(t\) is \(\vec{r}_B = v_1 t \hat{j}\). Boy A moves with velocity \(\vec{v}_A = v_{Ax}\hat{i} + v_{Ay}\hat{j}\). For A to catch B, their positions must be equal at time \(t\). So, \(a\hat{i} + \vec{v}_A t = v_1 t \hat{j}\). This implies \(v_{Ax} = -a/t\) and \(v_{Ay} = v_1\). The magnitude of A's velocity is \(v = |\vec{v}_A| = \sqrt{v_{Ax}^2 + v_{Ay}^2}\). So, \(v^2 = (-a/t)^2 + v_1^2\). Rearranging for \(t\): \(t^2 = \frac{a^2}{v^2 - v_1^2}\), hence \(t = \frac{a}{\sqrt{v^2 - v_1^2}}\).