Smallest at $t^*=\dfrac{l_1v_1+l_2v_2}{v_1^{2}+v_2^{2}}$, and the smallest gap is $s_{\min}=\dfrac{\lvert l_1v_2-l_2v_1\rvert}{\sqrt{v_1^{2}+v_2^{2}}}$.
(ii) Problem Statement
"Two particles, 1 and 2, move with constant velocities $v_1$ and $v_2$ along two mutually perpendicular straight lines toward the intersection point O. At the moment $t=0$ the particles were located at distances $l_1$ and $l_2$ from O. How soon will the distance between the particles become the smallest? What is it equal to?"
The trick: sit on particle 2. Then particle 1 drifts along a single straight line. The smallest gap is just the perpendicular distance from particle 2 to that line — no calculus needed.
(iii) The Big Idea in Plain Words
Two dots slide along two roads that cross at a point O. The roads meet at a right angle. Each dot moves at its own steady speed toward O.
Symbols: $l_1,l_2$ are the start gaps from O. $v_1,v_2$ are the steady speeds. $t^*$ is the time of closest approach. $s_{\min}$ is the smallest gap.
They usually miss each other — one reaches O before the other. So the gap shrinks, hits a smallest value, then grows again. We want that smallest value and when it happens.
(iv) The Solution Viewed in Each Frame
1. Ground Frame (stationary observer)
Why: put O at the origin, with each particle on its own axis.
Particle 1: position $x_1=l_1-v_1t$. Particle 2: position $y_2=l_2-v_2t$. Gap squared: $s^2=(l_1-v_1t)^2+(l_2-v_2t)^2$.
Why: the gap is smallest where its rate of change is zero.
Set $\dfrac{d(s^2)}{dt}=0$: $-v_1(l_1-v_1t)-v_2(l_2-v_2t)=0\Rightarrow t^*=\dfrac{l_1v_1+l_2v_2}{v_1^{2}+v_2^{2}}$.
Why: put $t^*$ back in.
$s_{\min}=\dfrac{\lvert l_1v_2-l_2v_1\rvert}{\sqrt{v_1^{2}+v_2^{2}}}$.
2. Particle-2 Rest Frame (relative velocity) — the clean route
Why: ride with particle 2. Then particle 1 moves in a straight line.
Its relative velocity is $\vec v_{12}=\vec v_1-\vec v_2$, with size $v_{12}=\sqrt{v_1^{2}+v_2^{2}}$ (the two legs are perpendicular).
Why: the closest the moving line gets to the fixed point is the perpendicular drop.
$s_{\min}=AP$ = perpendicular from particle 2 onto particle 1's straight path. Geometry gives $s_{\min}=\dfrac{\lvert l_1v_2-l_2v_1\rvert}{\sqrt{v_1^{2}+v_2^{2}}}$, reached at $t^*=\dfrac{l_1v_1+l_2v_2}{v_1^{2}+v_2^{2}}$.
3. Centre-of-Mass Frame
Does not apply. No masses are given and there is no interaction; a CM frame adds nothing here.
4. Medium Frame
Does not apply. There is no air, water or flowing medium.
- Changes: the picture — two dots on two roads (ground) versus one dot gliding past a fixed dot along a straight line (particle-2 frame).
- Same: the gap at every instant, so the same $t^*$ and the same $s_{\min}$.
(v) The Solution by Each Method
Method 1: Relative-Motion Method (fastest)
- Why: in particle 2's frame particle 1 goes straight.
Relative speed $v_{12}=\sqrt{v_1^2+v_2^2}$. - Why: closest approach is the perpendicular distance.
$s_{\min}=\dfrac{\lvert l_1v_2-l_2v_1\rvert}{\sqrt{v_1^2+v_2^2}}$ at $t^*=\dfrac{l_1v_1+l_2v_2}{v_1^2+v_2^2}$.
Method 2: Calculus Method
- Why: minimise the squared gap (no square root to differentiate).
$s^2=(l_1-v_1t)^2+(l_2-v_2t)^2$; set the derivative to zero. - Why: solve the linear equation for $t$.
$t^*=\dfrac{l_1v_1+l_2v_2}{v_1^2+v_2^2}$, then $s_{\min}=\dfrac{\lvert l_1v_2-l_2v_1\rvert}{\sqrt{v_1^2+v_2^2}}$.
Method 3: Graph Method
- Why: $s^2(t)$ is a parabola (a U-shape) in $t$.
Its lowest point is the vertex, at $t^*$. The bottom panel plots the gap $s(t)$ and marks the dip. - Why: read the dip height.
The bottom of the curve is $s_{\min}$.
Method 4: Force Method
Does not suit this problem. No forces act; both speeds are constant.
Method 5: Energy Method
Does not suit this problem. Speeds are fixed, so energy stays constant and tells us nothing about the gap.
Fastest: Method 1. Most reliable in an exam: Method 2. All give the same $t^*$ and $s_{\min}$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(vii) Common Exam Traps
- Assuming they collide at O. They meet at O only if $l_1/v_1=l_2/v_2$; otherwise one gets there first and $s_{\min}\gt0$.
- Using $v_1-v_2$ for the relative speed. The lines are perpendicular, so the relative speed is $\sqrt{v_1^2+v_2^2}$, not the plain difference.
- Minimising $s$ with the square root still in. Minimise $s^2$ instead — the algebra is far cleaner.
- Swapping the cross terms. The time uses $l_1v_1+l_2v_2$; the distance uses $l_1v_2-l_2v_1$. Do not mix them.
- Forgetting the absolute value on $s_{\min}$. Distance is never negative.
Verified: the live gap curve, both frames and all methods agree — $t^*=(l_1v_1+l_2v_2)/(v_1^2+v_2^2)$ and $s_{\min}=|l_1v_2-l_2v_1|/\sqrt{v_1^2+v_2^2}$, in SI units.