Worked Examples: Position–Time Graphs

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Module 3: Kinematics  ·  Worked Examples  ·  Prerequisites: K5, Foundation 7

Four problems that ask you to read a graph, build one, and judge whether a curve could be a motion at all.

Mahavakya

The height is the position. The slope is the velocity. Reading one when the question asked for the other is the commonest error in this topic.

Most graph questions ask you to pick a curve. None of these does. Two ask you to produce a full description, one asks you to construct the graph from words, and one asks whether four candidate curves could represent a motion at all.

1. Describe the motion completely

Problem

A particle’s position–time graph consists of three straight segments: from (0, 0) to (2, 6); horizontal to (5, 6); then down to (8, −3). Distances in metres, times in seconds.

Find the velocity in each segment, the total distance, the displacement, when it passes the origin, and say what the two corners hide.

A position-time graph in three straight segments: rising from the origin to six metres at two seconds, horizontal until five seconds, then falling to minus three metres at eight seconds, crossing the time axis at seven seconds. Both corners are circled.

Slope gives the velocity in each segment; the corners are where the rule changes.

0 to 2 s:   v = 6/2 = +3 m/s  —  forward, steady

2 to 5 s:   v = 0  —  at rest, six metres from the origin

5 to 8 s:   v = (−3 − 6)/3 = −3 m/s  —  backward, steady

Distance = 6 + 0 + 9 = 15 m.   Displacement = −3 − 0 = −3 m.

The origin. On the last segment, x = 6 − 3(t − 5), which vanishes at t = 7 s.

The corners are the part worth thinking about. At each one the velocity changes by 3 m/s with no time elapsing, which would need an infinite acceleration. Every segment is straight, so a = 0 everywhere else.

That does not make the graph impossible. It makes it an idealisation — a real transition takes a short interval during which a large acceleration acts, and the corner is what you get when that interval is squeezed to nothing. The position stays continuous throughout, which is what matters.

2. Build the graph from a description

Problem

A lift starts at the 4th floor and descends steadily to the ground in 8 s. It waits 6 s, then rises to the 6th floor in 15 s, decelerating over the last 5 s. Floors are 4 m apart.

Sketch x(t), and state what is continuous at each join.

A position-time graph for a lift: a straight fall from sixteen metres to the ground over eight seconds, a horizontal wait until fourteen seconds, a straight rise, then a curve flattening to reach twenty-four metres at twenty-nine seconds.

Position is continuous everywhere. Velocity is not, and that is where the corners are.

Descent. From x = 16 m to 0 in 8 s, so v = −2 m/s, a straight line.

Wait. Horizontal at x = 0, from t = 8 to 14 s.

Rise. Total climb 24 m in 15 s, of which 10 s at constant speed v and 5 s decelerating uniformly to rest. The decelerating stretch covers half what it would at full speed, so:

10v + 2.5v = 24

v = 24/12.5 = 1.92 m/s

t = 8 s — position continuous, velocity not. A corner.

t = 14 s — position continuous, velocity not. A corner.

t = 24 s — position and velocity continuous. Smooth.

Only the last join is smooth, because that is the only one where the problem describes the lift easing off rather than changing abruptly. The two corners are the idealisation again — a real lift takes a moment to start and stop, and the graph pretends it does not.

3. Which of these could be a motion?

Problem

Five curves are offered as position–time graphs. For each, say whether it could represent a motion — and if not, which test it fails.

Five panels of candidate position-time curves: a vertical segment, a step, a curve rising to a vertical tangent, a curve with a sharp corner, and a smooth sine wave. The first three are marked as failing a test, the fourth as an allowed idealisation, the fifth as valid.

Four questions, asked in order: single-valued? continuous? finite slope? smooth slope?

A — vertical segment. Impossible. A vertical line meets one instant twice, so the particle would be in two places at once. It fails the first test, and no later question needs asking.

B — step. Impossible. Single-valued, but the position jumps: the particle is in one place and then, with no time elapsing, somewhere else.

C — vertical tangent. Impossible. Single-valued and continuous, but the slope becomes unbounded, which is an infinite velocity. Note this is a different failure from A: a vertical segment fails the first test, a vertical tangent passes it and fails the third.

D — sharp corner. Allowed. This is the one that is not impossible. The velocity changes abruptly, which no real body does — but it is what a textbook draws for a ball bouncing off a wall, and the position stays continuous. It is a modelling choice, not a violation.

E — smooth sine. Valid. Single-valued, continuous, finite slope, smooth.

The curve that looks strangest is the only ordinary one. E describes a body oscillating back and forth, which is most of what mechanical systems do. A, B and C look far more like the graphs in a textbook and are all impossible.

So do not judge a curve by whether it resembles graphs you have seen. Ask the four questions.

The common mistake is calling D impossible along with the rest. Three of these are ruled out by physics; the fourth is ruled out only by a decision about how much detail to keep. Those are different objections, and treating them alike would forbid most of the graphs in this module.

4. Two particles on one graph

Problem

Two particles move along the same line with xA = t² − 6t + 12 and xB = −t² + 6t − 4, in metres and seconds.

(a) When do they meet? (b) When are their velocities equal? (c) Explain why these are different instants.

Two position-time curves crossing at two seconds and four seconds, with horizontal tangent lines drawn on both curves at three seconds showing that their slopes are equal there.

Equal velocity is not where the curves cross. It is where they run parallel.

(a) They meet where the positions agree:

t² − 6t + 12 = −t² + 6t − 4

2t² − 12t + 16 = 0  →  t² − 6t + 8 = 0

t = 2 s and t = 4 s

(b) Their velocities agree where the derivatives match:

vA = 2t − 6    vB = −2t + 6

2t − 6 = −2t + 6  →  4t = 12

t = 3 s, where both velocities are zero

(c) Crossing and matching are different questions. The curves cross where the heights agree — the particles are at the same place. Their velocities agree where the slopes agree — that is, where the curves run parallel, which here happens midway between the two meetings.

This is the Mahavakya in its sharpest form. Height and slope are independent readings of the same curve, and a question about one cannot be answered by looking at the other.

What the four have in common

None of them can be answered by matching the picture to a remembered shape. The first asks for a description, the second for a construction, the third for a judgement, the fourth for a distinction.

Three of the four turn on the same idea: a corner, a crossing, a tangent — each is a feature of the curve that means something specific, and means nothing at all if you read the wrong quantity off it.

Reading any position–time graph

  1. Height — where the body is.
  2. Slope — how fast, and which way.
  3. Curvature — whether that is changing.
  4. Corners and crossings — what each one means, before assuming it means anything.

Where these sit

Examples 1 and 2 are JEE Main standard. Examples 3 and 4 are Advanced — 3 for the admissibility judgement, 4 for the height-against-slope distinction. All four assume K5; example 2 also touches K10’s continuity hierarchy.

Prerequisites: K5, Foundation 7  ·  Working time: 35 min  ·  Level: JEE Main to Advanced

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