Question

Two particles A and B move constant velocities $v_1$ and $v_2$ along two mutually perpendicular straight lines towards the intersections point O. At moment t = 0, the particles were located at distance $d_1$ and $d_2$ from O respectively. Find the time, when they are nearest and also this shortest distance.

Answer 1

At time $t=t$, the positions are $A^{\prime }$ and $B^{\prime }$.

Now, $O{A}^{\prime }=d_1-v_{1}t$ and $O{B}^{\prime }=d_2-v_{2}t$

Separation is $l=\sqrt{O{A}^{\prime 2}+O{B}^{\prime 2}}=\sqrt{(d_1-v_{1}t{)}^2+(d_2-v_{2}t{)}^2}$

For shortest distance, $\frac{dl}{dt}=0$

$⟹\frac{-2v_1(d_1-v_{1}t)-2v_2(d_2-v_{2}t)}{2\sqrt{(d_1-v_{1}t{)}^2+(d_2-v_{2}t{)}^2}}=0$

$⟹(v_1{d}_1+v_2{d}_2)-t(v_1^2+v_2^2)=0$

$⟹t=\frac{v_1{d}_1+v_2{d}_2}{v_1^2+v_2^2}$ for least separation.

Now, separation at this time is-

$O{A}^{\prime }=d_1-v_{1}t=\frac{v_2(v_2{d}_1-v_1{d}_2)}{v_1^2+v_2^2}$

$O{B}^{\prime }=d_2-v_{2}t=\frac{-v_1(v_2{d}_1-v_1{d}_2)}{v_1^2+v_2^2}$

$l=\sqrt{O{A}^{\prime 2}+O{B}^{\prime 2}}=|v_2{d}_1-v_1{d}_2|$ as distance is always +ve.