ARCHETYPE: Chase, Catch & Meet

Around 450 BCE, the philosopher Zeno of Elea claimed that swift Achilles could never catch a tortoise with a head start: by the time Achilles reaches where the tortoise was, the tortoise has crawled a little further, forever. The paradox teased thinkers for two thousand years, and the modern resolution is almost insolently simple — write each runner's position as a function of time, set the two expressions equal, and solve. The infinitely many "catching-up stages" collapse into one linear equation with one finite answer. That single move, position functions plus a shared clock, is the master key to an entire family of examination problems. The family is everywhere. A police jeep accelerates after a thief who sped past a checkpoint. A man sprints for a bus that has just pulled away. Two trains close in on one another on the same line. Car B leaves two hours after car A and must catch it. CBSE and ISC set these as structured three-mark problems; JEE-Main and NEET compress them into two-minute MCQs; JEE-Advanced sharpens them into "find the minimum speed to catch" questions where the chase only just succeeds. Different costumes, same skeleton: two movers, one axis, one meeting condition. So this article treats the setup itself as the concept — the way the Atwood machine module treats pulleys. You will build the master pattern once, learn the relative-motion shortcut that halves the algebra, decode what the quadratic's two roots physically mean, and drill the four variants examiners recycle every year. After that, every chase on paper is fill-in-the-blanks.

Key takeaways

Every chase-and-meet problem is two position functions on one axis with one shared origin and one shared clock: $x_1(t)$ and $x_2(t)$ .
The whole physics of "meet", "catch", "overtake", "collide" is a single condition: $x_1(t) = x_2(t)$ — solve for $t$ , substitute back for the place.
The relative-motion shortcut collapses two movers into one: track the gap with relative velocity $v_{\text{rel}}$ and relative acceleration $a_{\text{rel}}$ ; the chase ends when the gap hits zero.
When the meeting equation is quadratic, its discriminant speaks: two roots = catch and re-pass, equal roots = barely catches (the minimum-speed case), no real roots = never meets.
Equal speeds is the moment of closest approach or maximum lead — never, by itself, the meeting.
Key idea: put both movers on one axis with one clock, and every chase question becomes "solve $x_1(t) = x_2(t)$ ".

The core insight

Two objects on the same line meet when their position functions, written from a common origin with a common clock, are equal — $x_1(t) = x_2(t)$ ; every chase, catch, overtake and head-on problem is that one equation, and the relative-motion form (gap shrinking at $v_{\text{rel}}$ , changing at $a_{\text{rel}}$ ) is the same equation with half the algebra.

Where this lives today

Rendezvous is the chase-and-meet archetype flown for real. When a cargo spacecraft docks with the International Space Station, guidance computers propagate two position functions and steer the relative state — gap and closing rate — to zero at contact, exactly the $g(t) \to 0$ formulation of this module (with orbital dynamics supplying fancier accelerations). Air-traffic control runs the same mathematics defensively: conflict-alert software continuously solves "will these position functions ever get too close?", which is the no-real-roots condition read as a safety guarantee. On the ground, adaptive cruise control and automatic emergency braking are chase problems solved thirty times a second: your car measures the gap and closing speed to the vehicle ahead by radar, computes whether the gap function will reach zero, and intervenes when the discriminant threatens to turn positive. And every navigation app's "you will catch the 6:40 ferry" is a delayed-start catch-up computation — Worked Example 1 with live traffic feeding the speeds.

The setup

Geometry first, because in an archetype the geometry is the physics. One straight road, taken as the $x$ -axis. Two movers — call them the leader (head start $d$ , velocity $u_1$ , acceleration $a_1$ ) and the chaser (starts at the origin, velocity $u_2$ , acceleration $a_2$ ). One origin, one positive direction, and — this is the part that saves marks — one clock, started at a single agreed event.

Figure

The chase axis — two movers, one origin, one clock, one gap

Each mover then carries its own uniformly-accelerated position function — the second kinematic equation from the Kinematic Equations module, promoted from "displacement" to "position" by adding the starting coordinate:

x_1(t) = d + u_1 t + \tfrac{1}{2}a_1 t^2, \qquad x_2(t) = u_2 t + \tfrac{1}{2}a_2 t^2 \qquad (1)

where: $d$ — the leader's head start (its position when the clock starts); $u_1, u_2$ — velocities at $t = 0$ ; $a_1, a_2$ — accelerations (zero for constant-speed movers). Read these as complete biographies: feed any $t$ into $x_1$ and it reports where the leader is, no storytelling required. All the drama of a chase — who is ahead, by how much, closing or opening — lives in the difference of these two functions, which is why the shortcut below works. Two conventions do the silent heavy lifting. First, head-on problems put the movers at opposite ends: give the second mover position $x_2(t) = L - u_2 t - \tfrac12 a_2 t^2$ if it starts at $x = L$ moving backward — its velocity enters with a minus sign on the shared axis. Second, delayed starts keep one clock: if the chaser leaves a time $t_0$ after the leader, the leader has been moving for $t$ but the chaser only for $(t - t_0)$ , so the chaser's function uses $(t - t_0)$ wherever $t$ appears. Never run two clocks.

The master solution pattern

Every question in the family is one of three sentences about the same pair of functions: when are they equal (meet/catch/overtake), when is their difference extreme (closest approach, maximum lead), or can they ever be equal (minimum speed to catch). The pattern: write equation (1) with correct signs, then impose the sentence.

Derivation AThe meeting condition and what its roots mean

We derive the general meeting time for two uniformly accelerated movers and — the exam-critical part — decode the physical meaning of the quadratic's two roots. The strategy: set the position functions equal, collect into a standard quadratic in $t$ , and interrogate the discriminant. Watch for the physical filter at the end: mathematics happily returns negative times and both intersections, and you must select the root the question is asking about.

Step 1. Impose the meeting sentence on equation (1): $x_2(t) = x_1(t)$ , i.e.

u_2 t + \tfrac12 a_2 t^2 = d + u_1 t + \tfrac12 a_1 t^2

Step 2. Collect everything on one side into standard form:

\tfrac12 (a_2 - a_1)\,t^2 + (u_2 - u_1)\,t - d = 0

Why: only the DIFFERENCES of the velocities and accelerations survive, plus the initial gap. The chase never cared about the individual motions — a first hint of the relative-motion shortcut.

Step 3. If $a_1 = a_2$ (both constant-speed, or equal accelerations), the quadratic degenerates to a linear equation with the single root

t = \frac{d}{u_2 - u_1}

Why: gap over closing speed — the most quoted formula in the family. It requires $u_2 > u_1$ : a slower chaser never closes a gap that shrinks at a constant rate.

Step 4. Otherwise solve the quadratic. The discriminant $D = (u_2-u_1)^2 + 2(a_2-a_1)d$ classifies the chase: $D > 0$ with two positive roots means the chaser catches, passes, and is later re-passed (the second root is the re-overtake, not an error); $D = 0$ means the chaser barely touches the leader at exactly one instant — the minimum-speed / just-catches case; $D < 0$ means the gap never closes.

Step 5. Physically filter the roots: discard negative times, and read the question — "when does he catch" wants the smaller positive root; "when does the bus escape again" wants the larger. Verified: limits — $d = 0$ gives $t = 0$ (they start together, trivially "meeting"); $a_1 = a_2$ and $u_1 = u_2$ gives no solution for $d \neq 0$ (identical motion preserves the gap forever). Worked Example 4's numbers land on this template: $t^2 - 12t + 20 = 0$ gives $t = 2$ and $t = 10$ s, catch then re-pass.

Go deeper · Derivation B · 1 minThe relative-motion shortcut — collapsing two movers into one

Derivation BThe relative-motion shortcut — collapsing two movers into one

We re-derive the meeting condition by riding along with the leader, which converts every two-body chase into a one-body problem and typically halves the algebra. The idea: only the gap matters, so track the gap directly. Its rate of change is the relative velocity and its curvature is the relative acceleration — the 1D case of the Relative Velocity & Reference Frames module. The thing to watch: choose ONE sign convention for "closing" and stay loyal to it.

Step 1. Define the gap $g(t) = x_1(t) - x_2(t)$ , the leader's lead. Subtracting the two lines of equation (1):

g(t) = d + (u_1 - u_2)\,t + \tfrac12 (a_1 - a_2)\,t^2

Why: the gap is itself a uniformly-accelerated "particle" with initial value $d$ , initial velocity $u_1 - u_2$ and acceleration $a_1 - a_2$ . Every kinematic equation you know applies to the gap directly.

Step 2. Rename in chase language: the chaser closes at relative velocity $v_{\text{rel}} = u_2 - u_1$ and relative acceleration $a_{\text{rel}} = a_2 - a_1$ . The gap obeys $g(t) = d - v_{\text{rel}}\,t - \tfrac12 a_{\text{rel}}\,t^2$ .

Step 3. The meeting sentence becomes one clean statement: $g(t) = 0$ . For constant speeds this is $d = v_{\text{rel}}\,t$ , i.e. time = gap ÷ closing speed, and for head-on motion the closing speed is $u_1 + u_2$ because the velocities carry opposite signs on the axis.

Step 4. The closest-approach sentence becomes: the gap is extremal when its rate of change vanishes — when the two speeds are momentarily equal. Feed that instant back into $g(t)$ for the minimum gap or maximum lead. Why: this is the origin of the most important non-obvious fact in the family — equal speeds is the turning point of the gap, NOT the meeting. A chaser still slower than the leader is still losing ground; the catch, if it comes, comes after the speeds cross. *Verified: the v²-over-2a check — a chaser at constant $v$ behind a leader accelerating from rest at $a$ has gap extremal at $t = v/a$ , where the gap has shrunk by $v^2/2a$ ; catching requires $d \le v^2/2a$ , reproducing the discriminant condition of the previous derivation exactly. Both routes agree, as they must.*

The pattern in table form — the four variants exams actually pose:

variant	setup on the axis	condition to impose	classic answer shape
catch-up (both constant speed)	chaser behind by $d$ , faster	$g = 0$	$t = \dfrac{d}{u_2 - u_1}$
head-on meeting	start $L$ apart, opposite velocities	$x_1 = x_2$	$t = \dfrac{L}{u_1 + u_2}$
police–thief (chaser accelerates from rest)	thief constant $v_t$ passes post; jeep starts, $a$	$\tfrac12 a t^2 = v_t\,t$	$t = \dfrac{2v_t}{a}$ ; jeep speed $= 2v_t$ at catch
bus chase (leader accelerates from rest)	runner constant $v$ , bus gap $d$ , accel $a$	$vt = d + \tfrac12 a t^2$	catches iff $v \ge \sqrt{2ad}$ ; equality = graze

Figure

Position–time picture of a chase — intersection is the catch, tangency is the closest call

Limiting cases

case	result	meaning
$u_2 \gg u_1$ , no accelerations	$t \to d/u_2$	overwhelming chaser: time is just gap over own speed
$u_2 \to u_1$ (constant speeds)	$t = \dfrac{d}{u_2-u_1} \to \infty$	matching speed freezes the gap — the chase never ends
$d \to 0$	$t \to 0$ (plus a second root if accelerations differ)	starting level: any speed edge catches instantly
head-on, one mover stopped	$t = L/u$	the moving one covers the whole separation alone
bus chase, $v = \sqrt{2ad}$ exactly	discriminant zero, one root $t = v/a$	grazes the bus at matched speeds — the minimum-speed catch
bus chase, $v < \sqrt{2ad}$	no real root; closest gap $= d - \dfrac{v^2}{2a}$	never catches; the shortfall is read off the gap function

Common trap"They meet when their speeds become equal"

Equal speeds is the moment the gap stops changing — the closest approach if the chaser was gaining, the maximum lead if it was losing — and in general nobody is beside anybody at that instant. Meeting is about equal positions, and speeds at the meeting moment are usually different (the accelerating police jeep passes the thief doing exactly twice the thief's speed). The confusion is fatal in the bus-chase family: at the just-catches minimum the runner reaches the bus precisely when their speeds match — position AND velocity equal simultaneously — and examiners exploit that special case to make you believe equal speeds is the general meeting test. It is the tangency condition, the boundary of the family, not its rule. Test yourself: in Worked Example 4 the speeds match at $t = 6$ s, while the catches happen at $t = 2$ s and $t = 10$ s.

Worked examples

1 · Board level — the delayed start (CBSE/ISC pattern, 3 marks)

Car A leaves a toll gate at a steady $40\ \mathrm{km/h}$ . Two hours later, car B leaves the same gate at a steady $60\ \mathrm{km/h}$ along the same road. When and where does B catch A?

Step 1. One origin (the gate), one clock — start it when B leaves. A has a head start $d = 40 \times 2 = 80\ \mathrm{km}$ .

Step 2. Positions: $x_A = 80 + 40t$ , $x_B = 60t$ (in km, t in hours). No accelerations, so the meeting is linear.

Step 3. Impose $x_B = x_A$ : $60t = 80 + 40t \Rightarrow 20t = 80 \Rightarrow t = \boxed{4\ \mathrm{h}}$ after B starts (6 h after A started). Shortcut check: gap ÷ closing speed $= 80/(60-40) = 4$ h ✓.

Step 4. Where: $x_B = 60 \times 4 = \boxed{240\ \mathrm{km}}$ from the gate. Cross-check with A: $80 + 40 \times 4 = 240$ km ✓.

2 · Board level — head-on trains (ISC/CBSE pattern)

Two stations lie $300\ \mathrm{km}$ apart. A passenger train leaves the first at $60\ \mathrm{km/h}$ while a goods train simultaneously leaves the second, heading toward it at $90\ \mathrm{km/h}$ . When and where do they meet?

Step 1. Shared axis from the first station. Passenger train: $x_1 = 60t$ . Goods train starts at $x = 300$ moving in the negative direction: $x_2 = 300 - 90t$ . The minus sign IS the head-on physics.

Step 2. Impose $x_1 = x_2$ : $60t = 300 - 90t \Rightarrow 150t = 300 \Rightarrow t = \boxed{2\ \mathrm{h}}$ . The closing speed is the sum $60 + 90 = 150\ \mathrm{km/h}$ — opposite velocities close the gap together.

Step 3. Meeting point: $x = 60 \times 2 = \boxed{120\ \mathrm{km}}$ from the first station (and $300 - 90\times2 = 120$ ✓).

Step 4. Classic add-on: a bird flying at $120\ \mathrm{km/h}$ back and forth between the approaching engines covers — no infinite series needed — simply speed × time $= 120 \times 2 = 240\ \mathrm{km}$ . The bird question is the meet question wearing a costume.

3 · Competitive level — police and thief (JEE-Main/NEET pattern, ~2.5 minutes)

A thief on a motorbike passes a stationary police jeep at a constant $54\ \mathrm{km/h}$ . At that very instant the jeep gives chase with constant acceleration $3.0\ \mathrm{m/s^2}$ . Find when and where the jeep catches the bike, and the jeep's speed at that moment.

Step 1. Convert units before anything: $54\ \mathrm{km/h} = 54 \times \dfrac{5}{18} = 15\ \mathrm{m/s}$ . Clock starts at the passing moment; both begin at the origin, so $d = 0$ .

Step 2. Positions: bike $x_t = 15t$ , jeep $x_p = \tfrac12 (3.0)t^2 = 1.5t^2$ . Impose $x_p = x_t$ : $1.5t^2 = 15t \Rightarrow t(1.5t - 15) = 0$ .

Step 3. The root $t = 0$ is the passing moment itself; the catch is the other root: $t = \boxed{10\ \mathrm{s}}$ , at $x = 15 \times 10 = \boxed{150\ \mathrm{m}}$ .

Step 4. Jeep's speed at the catch: $v = at = 3.0 \times 10 = \boxed{30\ \mathrm{m/s}}$ — exactly twice the thief's speed. Not a coincidence: catching from rest with equal start means equal average speeds, and a uniformly accelerating jeep averaging $15$ must finish at $30$ . Timing note: ~2.5 min. The factor-of-two ending is a reusable theorem — for any start-from-rest chase with $d=0$ , catch time $t = 2v_t/a$ and catch speed $= 2v_t$ . Quote it and this becomes a 40-second question.

4 · Competitive level — the bus chase, minimum speed (JEE-Advanced flavour, ~4 minutes)

A student stands $10\ \mathrm{m}$ behind a bus when it starts from rest with constant acceleration $1.0\ \mathrm{m/s^2}$ . (a) Running at a constant $6.0\ \mathrm{m/s}$ , when does she reach the bus? (b) What is the minimum constant speed with which she could just catch it?

Step 1. Gap function (relative-motion shortcut): the gap starts at $d = 10$ m, closes at $v_{\text{rel}} = 6.0 - t \cdot(1.0)$ ... better, write it directly: $g(t) = 10 + \tfrac12(1.0)t^2 - 6.0t = 0.5t^2 - 6t + 10$ .

Step 2. Catch when $g = 0$ : $t^2 - 12t + 20 = 0 \Rightarrow t = \dfrac{12 \pm \sqrt{144 - 80}}{2} = \dfrac{12 \pm 8}{2}$ , so $t = \boxed{2\ \mathrm{s}}$ (catches the door) — and $t = 10$ s is the moment the accelerating bus, if she never boarded, would sweep past her again. Two roots, two real events.

Step 3. Minimum speed: catching just barely means the discriminant vanishes. For speed $v$ : $0.5t^2 - vt + 10 = 0$ , discriminant $v^2 - 4(0.5)(10) = v^2 - 20 = 0$ , so $v_{\min} = \sqrt{20} = 2\sqrt5 \approx \boxed{4.5\ \mathrm{m/s}}$ . This is the template $v_{\min} = \sqrt{2ad} = \sqrt{2 \times 1.0 \times 10}$ ✓.

Step 4. At the graze, $t = v/a = 4.47$ s and the bus speed is $at = 4.47\ \mathrm{m/s}$ — equal to hers. The minimum-speed catch happens exactly at matched speeds: the tangency of the position–time curves, and the one case where "equal speeds" and "meeting" coincide. Timing note: ~4 min the first time, ~90 s once $v_{\min} = \sqrt{2ad}$ and the tangency picture are yours. On MCQs, check the discriminant before hunting roots — "does she catch it at all?" is often the actual question.

Exam traps

Two clocks. In delayed-start problems, writing $x_A = 40t$ and $x_B = 60t$ with each $t$ meaning "time since that car started" makes the positions incomparable. One clock: $x_A = 40t$ , $x_B = 60(t - 2)$ for $t \ge 2$ — or restart the clock at B's departure and bank A's head start as $d$ , as in Worked Example 1. Either discipline works; mixing them costs the full three marks.

Head-on velocities entered with the same sign. On one axis, movers approaching each other have oppositely-signed velocities; the closing speed is the sum of the speeds. Writing $x_2 = 300 + 90t$ for a train coming toward the origin sends it the wrong way and produces a nonsense meeting behind the second station.

Equal speeds read as the catch. Equal speeds is the gap's turning point — closest approach or maximum lead (see the warning box). Impose equal positions for meetings; use equal speeds only when the question says "closest", "maximum separation", or "just barely".

Dropping the second root. A quadratic meeting equation with two positive roots is telling a story: catch, then re-pass. "When does he catch it" wants the smaller root; discarding "the ugly root" mechanically loses the part (b) that JEE-Advanced loves to ask.

Forgetting km/h → m/s. Mixed units are deliberately planted: $54\ \mathrm{km/h}$ against $3\ \mathrm{m/s^2}$ must become $15\ \mathrm{m/s}$ (multiply by $5/18$ ) before any equation is written. A jeep "catching" the thief in $3.6$ s is the tell that someone skipped it.

Answering the time but not the place (or vice versa). "When and where" is two answers and usually two marks. Substitute your $t$ back into either position function — and into the other as a free cross-check.

Negative or zero roots discarded thoughtlessly. $t = 0$ in Example 3 was the passing event that started the clock — physical, just not the answer. A negative root means "the meeting already happened before the clock started" in the extrapolated motion; state that, don't just cross it out, when a question asks you to interpret.

Origins · 1 minRené Descartes and how this was discovered

René Descartes

1596–1650 · Touraine, the Dutch Republic & Stockholm

Descartes made "where is it, as a function of what?" a mathematical question. His La Géométrie (1637) — published as an appendix to the Discourse on Method — welded algebra to geometry: curves became equations, and the intersection of two curves became the simultaneous solution of two equations. That is precisely the machinery of this module. A chase is two graphs in the position–time plane; the catch is their intersection; the just-catches case is tangency — all Cartesian vocabulary. Legend says the coordinate idea came to him watching a fly wander the ceiling from his bed, wondering how to report its position with numbers. Whether or not the fly existed, every $x(t)$ you write, every meeting you find by setting two expressions equal, is Descartes' program running exactly as designed, four centuries on.

History

450 BCE

— Zeno of Elea poses Achilles and the tortoise: the swiftest runner can "never" catch a head start. The paradox stands unresolved for two millennia.

1637

— Descartes' La Géométrie turns curves into equations; meetings become intersections, and Zeno's infinite stages collapse into one solvable condition.

1638

— Galileo's Two New Sciences gives uniform and uniformly-accelerated motion their exact laws — the two position functions every chase problem uses.

1687

— Newton's Principia founds kinematics on the calculus of position with respect to a universal clock: one time variable for every body at once.

1896

— Walter Arnold earns history's first speeding conviction, caught by a policeman on a bicycle — a literal constant-acceleration chase-and-meet.

1966

— Gemini 8 completes the first orbital docking: a chase-and-meet problem solved at 7.8 km/s, with the same mathematics of closing gaps.

Going further

Texts. NCERT Physics Class 11, Ch. 3 (Motion in a Straight Line, relative-velocity section) · HC Verma, Concepts of Physics Vol. 1, Ch. 3 · Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 2 (the moving-bodies problem set).

Provenance. All results are standard uniform-acceleration kinematics. A verifier should recompute: Example 1 ( $t = 4$ h, $x = 240$ km by both functions); Example 2 ( $t = 2$ h, $120$ km, bird $240$ km); Example 3 ( $t = 10$ s, $x = 150$ m, jeep at $30\ \mathrm{m/s} = 2 \times 15$ ); Example 4 ( $t^2 - 12t + 20 = 0 \Rightarrow t = 2, 10$ s; $v_{\min} = \sqrt{2 \times 1.0 \times 10} = 4.47\ \mathrm{m/s}$ ; graze at $t = 4.47$ s with matched speeds); the closest-gap formula $d - v^2/2a$ in the limiting-cases table.

Prerequisites. Kinematic Equations for Uniform Acceleration · Acceleration. Leads to. Trains Crossing & Overtaking · River–Boat Crossing · Graph Reading at Speed — with the frame-changing generalization in Relative Velocity & Reference Frames.

Do I start the clock when the first object moves or when the second one does?

Either — but exactly one. Both conventions appear in textbook solutions, which is why this confuses everyone at 2am. Starting the clock at the later event (Worked Example 1) is usually cleaner: the earlier mover's progress so far becomes a plain head start $d$ , and both functions then use the same $t$ with no $(t - t_0)$ bookkeeping. If you instead start at the first event, write the latecomer's position with $(t - t_0)$ and remember it only exists for $t \ge t_0$ . What you must never do is give each mover its own personal $t$ .

How do I know whether to expect one meeting or two?

Count the curvatures. Two constant-speed movers have straight position–time graphs: at most one intersection. The moment one mover accelerates, its graph curves, and a line can cut a parabola twice — catch and re-pass, like Worked Example 4's $t = 2$ s and $t = 10$ s. The discriminant of the meeting quadratic is the formal test: positive means two meetings, zero means a single grazing touch, negative means never. Asking "which root does the question want?" before solving saves you from reporting the re-pass as the catch.

Why is the closing speed the SUM of the speeds for head-on motion? It feels like double counting.

Because both movers eat into the same gap simultaneously. In one second, a 60 km/h train removes 60/3600 km from the gap at one end while the 90 km/h train removes 90/3600 km at the other: total shrinkage 150/3600 km, no double counting — two workers, one trench. Formally, on a shared axis the velocities are $+60$ and $-90$ , and the gap's rate of change is their difference, $60 - (-90) = 150$ km/h. The sum appears because subtracting a negative adds.

In the police-thief problem, why does the jeep end up at exactly twice the thief's speed? Is that always true?

It is a theorem for the specific — and most common — setup: both start from the same point at the same moment, the leader at constant $v_t$ , the chaser from rest with constant acceleration. Over the chase, equal positions in equal time force equal average velocities. The thief's average is $v_t$ ; the jeep's average, accelerating uniformly from rest, is half its final speed. Half the final speed $= v_t$ means final speed $= 2v_t$ , independent of the acceleration's value — a larger $a$ shortens the chase but still delivers the factor two. Give the leader a head start and the tidy factor breaks.

Can I always use the relative-velocity shortcut instead of two position functions?

For finding when — yes, and it is usually faster: gap, closing rate, closing acceleration, solve $g(t) = 0$ . Its blind spot is where: the shortcut lives entirely in the gap and forgets the ground. To report the meeting place you must return to an actual position function and substitute the meeting time. A good exam rhythm: shortcut for $t$ , one line of $x_2(t)$ for the place, the other position function as a free cross-check.

What if BOTH objects accelerate?

Nothing new — the machinery absorbs it. The meeting equation keeps the form of the master derivation with $a_{\text{rel}} = a_2 - a_1$ ; if the accelerations happen to be equal, the $t^2$ terms cancel and the problem collapses back to the linear gap-over-closing-speed case (equal accelerations are invisible to the gap). JEE-Advanced enjoys the special case where the chaser both starts slower and accelerates harder: the gap first grows, peaks when the speeds match, then shrinks to the catch — three phases, one quadratic.

Are these problems connected to the graphs questions on my paper?

Deeply — the second figure is the whole story. On a position–time plot, meetings are intersections, speeds are slopes, closest approach is where the two tangents run parallel, and the just-catches case is tangency of the two curves. Many MCQs are solved fastest by sketching the two graphs and reading the geometry, no algebra at all: the Velocity–Time Graphs module and the Graph Reading at Speed technique module drill exactly that reflex. If a chase MCQ offers graphs among the options, geometry first, algebra second.