WE-5 — Worked Examples: Frames, Journeys and Motion in Stages

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Module 3: Kinematics  ·  Worked Examples  ·  Prerequisites: K1, K7, K10

Five problems about the two things a description of motion needs before any equation: where you are measuring from, and when the rule changes.

Mahavakya

Move the origin and every position changes. Displacement does not. What survives a change of description is what was physical to begin with.

The first two examples ask what a change of origin does and does not alter. The third solves one problem twice, from two frames. The last two follow a body whose governing equation changes partway through — where the only thing carrying across the boundary is the state.

1. Two observers, one journey

Problem

A particle moves along a straight track. Measured from origin O1, it starts at xi = −15 m and ends at xf = +25 m — but partway through it reaches xm = +35 m before coming back.

Find the displacement and the distance from O1. Then repeat from an origin O2 sitting at +10 m on the first scale, and say which quantities changed.

From O1. Displacement cares only about the endpoints:

Δx = 25 − (−15) = +40 m

s = |35 − (−15)| + |25 − 35| = 50 + 10 = 60 m

From O2. The new origin is at +10, so every reading drops by 10: x′ = x − 10.

xi′ = −25 m    xm′ = +25 m    xf′ = +15 m

Δx′ = 15 − (−25) = +40 m

s′ = |25 − (−25)| + |15 − 25| = 50 + 10 = 60 m

All three positions changed. Neither the displacement nor the distance did. That is not a coincidence of these numbers — both are built from differences of positions, and a constant subtracted from every reading cancels in every difference.

So the origin is a choice, and quantities that depend on it are partly about the observer. The ones that survive the shift are the ones describing the particle.

Note what makes distance larger than displacement here. The particle overshoots to +35 and comes back — 60 m against 40 m. Without that reversal the two would be equal, and no origin shift would have revealed anything.

2. Three and a half laps

Problem

A runner completes 3.5 laps of a circular track of radius 50 m.

Find the distance and the displacement, form their ratio, and state exactly when the two can be equal.

s = 3.5 × 2π(50) = 350π ≈ 1100 m

Half a lap past the start leaves the runner diametrically opposite it:

|Δr| = 2R = 100 m

s / |Δr| = 350π/100 = 3.5π ≈ 11.0

The ratio can be made as large as you like — run more laps and it grows without limit. It can never fall below 1, because the straight line between two points is the shortest path between them.

Equality needs a straight line travelled in one direction only. Not a straight path with a reversal in it; not a curve, however gentle. Any deviation and any turning back makes s strictly larger.

The trap is the half lap. A student who takes 3.5 laps and reasons “3 whole laps cancel, so half a lap remains” is right — but the displacement of half a lap is the diameter, 2R, not the arc πR. Draw it.

3. One pursuit, two frames

Problem

A cyclist travelling at 15 m/s sets off after a scooter that is 100 m ahead and moving at 10 m/s, both steady and in the same direction.

Find when and where the cyclist catches the scooter — twice, by two different methods.

Method 1 — ground frame. Put the origin at the cyclist’s start, positive in the direction of travel:

xc = 15t     xs = 100 + 10t

15t = 100 + 10t  →  5t = 100

t = 20 s, at x = 300 m

Method 2 — the scooter’s frame. Sit on the scooter. It is at rest, and the cyclist approaches at

vcs = 15 − 10 = 5 m/s

closing 100 m at 5 m/s  →  t = 100/5 = 20 s

The second method is one line, and it is the same physics. Nothing was assumed that the first method did not also assume — the frames differ, the answer does not.

But notice what the second method does not give you: the position. In the scooter’s frame the meeting happens at the origin, and recovering x = 300 m needs a return to the ground. Choosing a frame is choosing which questions become easy, and which need translating back.

4. A journey that turns around

Problem

A body starts from rest and accelerates at 4 m/s² for 5 s. It then decelerates at 6 m/s² and is observed until t = 10 s.

Find when it reverses, where it is at t = 10 s, and the total distance travelled.

Stage 1 ends with the state that starts stage 2:

v(5) = 4(5) = 20 m/s     x(5) = ½(4)(25) = 50 m

Stage 2 begins at 20 m/s, not at rest. Writing τ = t − 5:

v = 20 − 6τ = 0  →  τ = 10/3, so t = 25/3 ≈ 8.33 s

at the turn: x = 50 + 20(10/3) − 3(10/3)² = 250/3 ≈ 83.3 m

at t = 10 s: x = 50 + 20(5) − 3(25) = 75 m

Distance must be split at the reversal:

s = 83.3 + (83.3 − 75) = 91.7 m,   while the displacement is only 75 m

Two boundaries, two different kinds. At t = 5 s the equation changes but the motion does not — the body carries 20 m/s across it. At t = 8.33 s the equation is unchanged but the motion reverses. Only the second one splits the distance calculation.

5. Three stages and a quadratic

Problem

An automated train covers 2400 m between two stations. It accelerates from rest at 1.5 m/s², cruises at its top speed for 60 s, then brakes at 2.0 m/s² to a stop.

Find the top speed, the duration of each stage, and the total journey time.

Express every stage in the one unknown. Both end stages start or finish at rest, so v² = 2as gives their distances directly:

s1 = v²/3     s2 = 60v     s3 = v²/4

Their sum is the distance between the stations:

v²/3 + 60v + v²/4 = 2400

7v² + 720v − 28800 = 0

v = 30.8 m/s  —  the negative root is rejected

t1 = 30.8/1.5 = 20.5 s    t2 = 60 s    t3 = 30.8/2.0 = 15.4 s    T = 95.9 s

The quadratic is not an accident. Two of the three stages have a distance proportional to v², and one proportional to v. Any three-stage profile of this shape produces a quadratic in the top speed, whatever the numbers.

And the answer is not round. That is the honest outcome here, and it is worth meeting — a student who has only ever seen problems engineered to give whole numbers will distrust a correct answer of 30.8 m/s.

What the five have in common

Every one of them turned on a decision made before any equation: where to put the origin, which frame to sit in, where one stage ends and the next begins.

None of those decisions is forced by the physics. Move the origin and the arithmetic changes completely while the motion does not. Sit on the scooter and a two-line problem becomes a one-line one. Draw the stage boundary in the wrong place and every number after it is wrong.

Which is why the quantities that survive those choices — displacement, distance, the relative velocity — are the ones worth trusting. They describe the particle. The rest describes us.

Carried forward

  • Shifting the origin changes every position and no difference between positions.
  • s ≥ |Δr| always, with equality only for straight, unreversed motion.
  • A moving frame can turn two equations into one — but positions must be translated back.
  • Neither the clock nor the velocity resets at a stage boundary.
  • A reversal splits a distance calculation. A change of equation does not.

Prerequisites: K1, K7, K10  ·  Working time: 35 min  ·  Level: JEE Main, with one Advanced

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