Question

Two particles A and B move with velocities v, and v, respectively along the x & y
axis. The initial separation between them is 'd' as shown in the fig. Find the least
distance between them during their motion.
>
А
X
d
V
B
dy
d.v?
(A) vi+v?
(B) vi+v
d.v.
d.v,
(C)
(D)
ſvi + v

Answer 1

Correct Question:-

Two particles A and B move with velocities $\sf{v_A}$ and $\sf{v_B}$ respectively along x & y axis. The initial separation between them is $\sf{d.}$ Find the least distance between them during their motion.

Solution:-

After a time $\sf{\Delta t,}$ the two particles moves by the distance $\sf{v_1\,\Delta t}$ and $\sf{v_2\,\Delta t}$ respectively.

Their positions will be at a distance $\sf{v_1\,\Delta t}$ and $\sf{d-v_2\,\Delta t}$ respectively as in the figure.

So the new separation is,

$\sf{\longrightarrow d'=\sqrt{(v_1\,\Delta t)^2+(d-v_2\,\Delta t)^2}}$

$\sf{\longrightarrow d'=\sqrt{(v_1)^2(\Delta t)^2+d^2-2dv_2\,\Delta t+(v_2)^2(\Delta t)^2}}$

$\sf{\longrightarrow d'=\sqrt{\big((v_1)^2+(v_2)^2\big)(\Delta t)^2-2dv_2\,\Delta t+d^2}\quad\quad\dots(1)}$

For the minimum separation, its derivative with respect to time should be zero.

$\sf{\longrightarrow\dfrac{dd'}{d(\Delta t)}=0}$

$\sf{\longrightarrow \dfrac{d}{d(\Delta t)}\left(\sqrt{\big((v_1)^2+(v_2)^2\big)(\Delta t)^2-2dv_2\,\Delta t-d}\right)=0}$

$\sf{\longrightarrow\dfrac{2\big((v_1)^2+(v_2)^2\big)\Delta t-2dv_2}{2\sqrt{\big((v_1)^2+(v_2)^2\big)(\Delta t)^2-2dv_2\,\Delta t-d}}=0}$

$\sf{\longrightarrow2\big((v_1)^2+(v_2)^2\big)\Delta t-2dv_2=0}$

$\sf{\longrightarrow\Delta t=\dfrac{dv_2}{(v_1)^2+(v_2)^2}}$

Then (1) becomes,

$\sf{\longrightarrow d'=\sqrt{\big((v_1)^2+(v_2)^2\big)\left(\dfrac{dv_2}{(v_1)^2+(v_2)^2}\right)^2-2dv_2\cdot\dfrac{dv_2}{(v_1)^2+(v_2)^2}+d^2}}$

$\sf{\longrightarrow d'=\sqrt{\dfrac{d^2(v_2)^2}{(v_1)^2+(v_2)^2}-\dfrac{2d^2(v_2)^2}{(v_1)^2+(v_2)^2}+\dfrac{d^2\big((v_1)^2+(v_2)^2\big)}{(v_1)^2+(v_2)^2}}}$

$\sf{\longrightarrow d'=\sqrt{\dfrac{d^2(v_2)^2-2d^2(v_2)^2+d^2(v_1)^2+d^2(v_2)^2}{(v_1)^2+(v_2)^2}}}$

$\sf{\longrightarrow d'=\sqrt{\dfrac{d^2(v_1)^2}{(v_1)^2+(v_2)^2}}}$

$\sf{\longrightarrow\underline{\underline{d'=\dfrac{dv_1}{\sqrt{(v_1)^2+(v_2)^2}}}}}$