Question bank/JEE Main/Physics

Two particle A and B are located in x-y plane at points (0, 0)
and (0, 4 m). They simultaneously start moving with velocities.
${\overset{\rightarrow}{v}}_{A} = 3\overset{\hat{}}{j}m/s$and
${\overset{\rightarrow}{v}}_{B} = 4\overset{\hat{}}{i}m/s$. The shortest
distance between them will be

Azero

Show answer and solution▾

Assuming B to be at rest, A will move with velocity
${\overset{\rightarrow}{v}}_{AB}$ in the direction shown in figure. The
distance between them will first decrease from A to Cand then increase
beyond C.
[IMAGE] Minimum distance between them is BC which is equal to
$4\ sin53^{\circ} = 16/\mathbf{5}\mathbf{m}$

Source: Grass-G / P-05 Physics paper 10 October.docx