Question

Two persons $A$ and $B$ are located X-Y plane at the points $(0,0)$ and $(0,10)$ respectively. (The distances are measured in MKS units). At time $t=0$ they start moving simultaneously with velocities $\overrightarrow{v_A}=2\hat{j}{ms}^{-1}$ and $\overrightarrow{v_B}=2\hat{i}{ms}^{-1}$ respectively. Find the time after which $A$ and $B$ are at their closest distance.

Answer 1

$x=V_{A}t=2t\stackrel{\wedge }{i}y=2t\stackrel{\wedge }{j}$

distance between

A and B $\parallel$ S

$s=\sqrt{y^2+{(10-x)}^2}=\sqrt{4t^2+{(10-2t)}^2}{s}^2={4t}^2+{4t}^2-4t+100=8t^2-4t+100$

For minimum $S.\frac{ds}{dt}=0t=\frac{4}{16}=\frac{1}{4}sec$