Irodov 1.16: Closest Approach Along Two Perpendicular Rails

Problem Irodov 1.16

Two particles move at constant speeds $v_1$ and $v_2$ , each along its own straight line, and the two lines are mutually perpendicular and cross at a point $O$ . Both particles head toward $O$ along their own line. At time $t=0$ , particle 1 is a distance $l_1$ from $O$ and particle 2 is a distance $l_2$ from $O$ . Find the time $t_m$ at which the distance between the two particles is smallest, and find that smallest distance $l_{\min}$ .

The starting configuration at t=0: particle 1 a distance l_1 out along x moving at v_1, particle 2 a distance l_2 out along y moving at v_2, both heading toward O. The arrows make v_2 the faster. — The starting configuration at $t=0$ : particle 1 a distance $l_1$ out along $x$ moving at $v_1$ , particle 2 a distance $l_2$ out along $y$ moving at $v_2$ , both heading toward $O$ . The arrows make $v_2$ the faster.

Given

$l_1,\ l_2$
$v_1,\ v_2$
the two lines mutually perpendicular, meeting at $O$

Find

the instant

t_m

of closest approach, and the minimum separation

l_{\min}

Solution

Step 1Each particle’s position along its own line

Put the $x$ axis along particle 1’s line and the $y$ axis along particle 2’s line, both oriented outward from $O$ . Each particle moves at constant speed toward $O$ , so its coordinate decreases linearly from its starting distance.

\begin{aligned} x_1(t) &= l_1-v_1 t \\ y_2(t) &= l_2-v_2 t \end{aligned} \tag{1}

Step 2The squared separation as a function of time

The two particles sit on perpendicular axes through $O$ , so their separation is the hypotenuse of the two coordinates (1); squaring avoids carrying a square root through the minimization.

L^2(t) = x_1(t)^2+y_2(t)^2 = (l_1-v_1 t)^2+(l_2-v_2 t)^2 \tag{2}

Step 3Minimizing the separation

Differentiating (2) with respect to $t$ and setting the result to zero locates the closest approach.

\begin{align*}\dfrac{d(L^2)}{dt} &= -2v_1(l_1-v_1 t)-2v_2(l_2-v_2 t) \\ 0 &= -v_1 l_1-v_2 l_2+(v_1^2+v_2^2)t\end{align*}

(v_1^2+v_2^2)\,t_m = v_1 l_1+v_2 l_2

(3)

t_m = \dfrac{v_1 l_1+v_2 l_2}{v_1^2+v_2^2}

(4)

Step 4The minimum separation itself

The same scene at t=t_m, with the two positions marked and the dashed segment l_ between them. Because v_2 is the faster here, particle 2 has already run past O while particle 1 has not. — The same scene at $t=t_m$ , with the two positions marked and the dashed segment $l_{\min}$ between them. Because $v_2$ is the faster here, particle 2 has already run past $O$ while particle 1 has not.

Substituting (3)’s $t_m$ into the coordinates (1), each numerator collects into a single term proportional to $l_1 v_2-l_2 v_1$ .

\begin{align*}x_1(t_m) &= l_1-v_1\cdot\dfrac{v_1 l_1+v_2 l_2}{v_1^2+v_2^2} = \dfrac{v_2(l_1 v_2-l_2 v_1)}{v_1^2+v_2^2} \\ y_2(t_m) &= l_2-v_2\cdot\dfrac{v_1 l_1+v_2 l_2}{v_1^2+v_2^2} = \dfrac{-v_1(l_1 v_2-l_2 v_1)}{v_1^2+v_2^2}\end{align*}

Squaring and adding these two, the common factor $(l_1v_2-l_2v_1)^2$ leaves only $v_1^2+v_2^2$ behind, which cancels one power of the denominator.

\begin{align*}L^2(t_m) &= x_1(t_m)^2+y_2(t_m)^2 \\ &= \dfrac{(l_1v_2-l_2v_1)^2\left(v_1^2+v_2^2\right)}{(v_1^2+v_2^2)^2} \\ &= \dfrac{(l_1v_2-l_2v_1)^2}{v_1^2+v_2^2}\end{align*}

l_{\min} = \dfrac{|l_1 v_2-l_2 v_1|}{\sqrt{v_1^2+v_2^2}}

(5)

Check

If the two particles are identical in every respect, $v_1=v_2=v$ and $l_1=l_2=l$ , Step 3 gives $t_m=2vl/(2v^2)=l/v$ , exactly the instant both reach $O$ together, and Step 4 gives $l_{\min}=|lv-lv|/(v\sqrt2)=0$ , consistent with them meeting exactly at the intersection. The second derivative of $L^2$ is the constant $2(v_1^2+v_2^2)>0$ , confirming the stationary point is a minimum and not a maximum.

Remark

Nothing in the algebra required either particle to still be short of $O$ at $t_m$ : in the figure $v_2>v_1$ , so particle 2 has already run through the intersection and lies on the far side of it by the time of closest approach. The formula for $t_m$ only ever asks where the separation is smallest, not whether that happens before or after either particle’s own crossing, which is why it does not need a case split.