Question

Two particles, $1$ and $2$ move with constant velocities $v_1$ and $v_2$ along two mutually perpendicular straight lines towards the intersection point $O$. At the moment $t=0$ the particles were located at the distance $l_1$ and $l_2$ from the point $O$. How soon will the distance between the particles become the smallest? What is it equal to?

Answer 1

The situation is shown of figure. After time $t$ let separation between them is $x$, then
$x^2=(l_1-v_{1}t{)}^2+(l_2-v_{2}t{)}^2.....\left(i\right)$
$x$ to be minimum, $\frac{dx}{dt}=0$
Differentiating equation (i) w.r.t time
we have $\frac{d{x}^2}{dt}=\frac{d}{dt}\left[\right(l_1-v_{1}t{)}^2+(l_2-v_{2}r{)}^2]$
$2x\frac{dx}{dt}=2(l_1-v_{1}t)\times (-v_1)+2(l_1-v_{2}t)\times (-v_2)$
or $0=(l_1-v_{1}t)(v_1+(l_2-v_{2}t)v_2=(l_1{v}_1-l_2{v}_2)-t(v_1^2+v_2^2)$
which gives $t=\left(\frac{l_1{v}_1+l_2{v}_2}{v_1^2+v_2^2}\right)$
Now $x_{min}=\sqrt{(l_1-v_{1}t{)}^2+(l_2-v_{2}t{)}^2}.....\left(ii\right)$
Substituting value of $t$ in equation (ii), we get
$x_{min}=\sqrt{{[l_1-v_1\left(\frac{l_1{v}_1+l_2{v}_2}{v_1^2+v_2^2}\right)]}^2+{[l_1-v_2\left(\frac{l_1{v}_1-l_2{v}_2}{v_1^2+v_2^2}\right)]}^2}$
After solving, we get $x_{min}=\left(\frac{l_1{v}_2-l_2{v}_1}{v_1^2+v_2^2}\right)$.