$t^*=\dfrac{lv}{v^{2}-u^{2}}$
(ii) Problem Statement
"Point A moves uniformly with velocity $v$ so that the vector $\vec v$ is continually 'aimed' at point B which in its turn moves rectilinearly and uniformly with velocity $u \lt v$. At the initial moment of time $\vec v\perp\vec u$ and the points are separated by a distance $l$. How soon will the points converge?"
Reference check (important): the supplied solution writes the closing rate $-\dfrac{ds}{dt}=v-u\cos\alpha$ and the x-condition $uT=\displaystyle\int v\cos\alpha\,dt$, then prints $T=\dfrac{ul}{v^{2}-u^{2}}$. That last line is a transcription slip. Carrying the book's own two equations through, $l=vT-u\!\int\!\cos\alpha\,dt=vT-u\cdot\tfrac{uT}{v}=\tfrac{T(v^{2}-u^{2})}{v}$, so $\boxed{T=\dfrac{lv}{v^{2}-u^{2}}}$. This agrees with the official Irodov answer key ($t=lv/(v^2-u^2)$). This page uses the corrected $lv/(v^2-u^2)$ everywhere, and the live simulation (numerical pursuit integration) confirms it.
(iii) The Big Idea in Plain Words
A hunter A always points straight at a target B and runs at a fixed speed $v$. The target B walks in a straight line at a slower speed $u$. Because A keeps turning to face B, A's path is a curve, not a straight line.
Symbols: $l$ is the start gap. $v$ is A's speed. $u$ is B's speed. $s$ is the gap at time $t$. $\alpha$ is the angle between B's velocity $\vec u$ and the line from A to B. $t^*$ is the catch time.
Two simple facts solve it. First, the gap shrinks at rate $v-u\cos\alpha$. Second, A and B start level across the direction of $\vec u$ (because $\vec v\perp\vec u$), so to meet they must cover the same distance along $\vec u$. Add these two and the messy angle $\alpha$ cancels out.
(iv) The Solution Viewed in Each Frame
1. Ground Frame (stationary observer)
Why: track the gap $s$ between A and B.
A runs straight at B, so A closes the gap at the full $v$. B moves at angle $\alpha$ to the line of sight, so B's part along that line is $u\cos\alpha$, which opens the gap. Net: $-\dfrac{ds}{dt}=v-u\cos\alpha$.
Why: get a second equation that kills $\alpha$.
At the start $\vec v\perp\vec u$, so A and B sit level in the $\vec u$ direction. To meet, both must travel the same distance along $\vec u$. B travels $uT$. A travels $\int v\cos\alpha\,dt$. So $uT=\int v\cos\alpha\,dt$.
Why: combine. Integrate the first equation over the whole chase.
$l=\displaystyle\int_0^{T}(v-u\cos\alpha)\,dt=vT-u\!\int_0^{T}\!\cos\alpha\,dt=vT-u\cdot\dfrac{uT}{v}$. So $l=\dfrac{T(v^{2}-u^{2})}{v}$ and $t^*=\dfrac{lv}{v^{2}-u^{2}}$.
2. Frame Riding with B (B at rest)
Why: sit on B. Then B is a fixed post and A spirals in.
In this frame A's velocity is $\vec v-\vec u$. Its component straight toward B is $v-u\cos\alpha$ — exactly the closing rate. This is uniform, non-accelerating motion of B, so no pseudo-force is needed. The catch time is identical.
Why: the same cancellation, read as geometry.
A's displacement toward B over the chase must equal the start gap plus how much B "ran out from under" A along $\vec u$. That bookkeeping is the pair of equations above, giving the same $t^*=\dfrac{lv}{v^{2}-u^{2}}$.
3. Line-of-Sight (rotating) Frame
Why: spin the axes so one axis always lies along AB.
Along the line: the gap changes at $v-u\cos\alpha$. Across the line: B's sideways part $u\sin\alpha$ makes the line of sight rotate. We never need the rotation to get the time — integrating the along-line rate with the level-start condition already fixes $t^*=\dfrac{lv}{v^{2}-u^{2}}$. The rotating axes only explain why the path curves.
4. Medium Frame
Does not apply. There is no air, water or flowing medium in this problem.
- Changes: A's path — a curved pursuit trail on the ground, a shorter in-curling path when you ride with B.
- Same: the catch time $t^*=\dfrac{lv}{v^{2}-u^{2}}$, and the fact that A and B meet having covered equal distance $uT$ along $\vec u$.
(v) The Solution by Each Method
Method 1: Two-Equation (Relative / Projection) Method
- Why: the gap closes at the net of A's approach and B's retreat along the line.
$l=\displaystyle\int_0^T(v-u\cos\alpha)\,dt$. - Why: a level start across $\vec u$ forces equal travel along $\vec u$.
$uT=\displaystyle\int_0^T v\cos\alpha\,dt\;\Rightarrow\;\int_0^T\cos\alpha\,dt=\dfrac{uT}{v}$. - Why: substitute and the integral vanishes.
$l=vT-u\cdot\dfrac{uT}{v}\Rightarrow t^*=\dfrac{lv}{v^{2}-u^{2}}$.
Method 2: Force Method
Does not suit this problem. No forces are given; both speeds are fixed by the rule of motion, not by dynamics.
Method 3: Energy Method
Does not suit this problem. Both speeds are constant, so kinetic energy never changes — there is nothing to balance.
Method 4: Graph Method
- Why: plot the gap $s$ against time $t$.
Unlike a steady chase, $-ds/dt=v-u\cos\alpha$ is not constant, so $s(t)$ is a gentle curve: it starts dropping at $v$ (since $\alpha=90^\circ$, $\cos\alpha=0$) and steepens toward $v+u$ as A swings behind B. - Why: the catch is where the curve meets $s=0$.
For $l=6$ m, $v=2$ m/s, $u=1$ m/s it reaches zero at $t^*=\dfrac{6\cdot2}{4-1}=4$ s. The bottom graph in the simulation draws this curve and marks $t^*$.
Method 5: Calculus Method (the book's route)
- Why: write the two integral conditions.
$-\displaystyle\int_l^0 ds=\int_0^T(u\cos\alpha\text{ flipped})$… cleanly: $l=\displaystyle\int_0^T(v-u\cos\alpha)dt$ and $uT=\displaystyle\int_0^T v\cos\alpha\,dt$. - Why: eliminate $\displaystyle\int\cos\alpha\,dt$ between them.
From the second, $\int\cos\alpha\,dt=uT/v$. Put into the first: $l=vT-u^2T/v$. - Why: solve for $T$.
$l=\dfrac{T(v^{2}-u^{2})}{v}\Rightarrow t^*=\dfrac{lv}{v^{2}-u^{2}}$. If $u\ge v$ the denominator is $\le0$: A can never catch B.
Fastest: Method 1. Most reliable in an exam: Method 5 (it is the book's own derivation). All give $t^*=\dfrac{lv}{v^{2}-u^{2}}$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(vii) Common Exam Traps
- Copying $T=ul/(v^2-u^2)$ from a web solution. The numerator is $lv$, not $ul$ — check it by units and by the $u\to0$ limit: with $u=0$, A should take $l/v$, and $lv/(v^2-0)=l/v$. The wrong form gives $0$.
- Assuming the gap shrinks at a constant rate. It starts at $v$ (because $\alpha=90^\circ$) and grows toward $v+u$; the $s$–$t$ graph is curved.
- Using $t=l/(v-u)$ (a head-on chase). Here B moves sideways at the start, not away, so that formula is wrong.
- Forgetting the condition $u \lt v$. If $u\ge v$ the points never converge.
- Missing the "equal distance along $\vec u$" condition — it is the second equation that makes $\alpha$ disappear.
Verified: the live numerical pursuit, all three frames and the methods give $t^*=lv/(v^2-u^2)$, with $l$ in m, $v,u$ in m/s and $t$ in s.