Module 1: Mathematical Foundations · Practice · assumes Foundation 7
Three worked examples where the graph is the question — not an illustration of it.
What this set is for
Every earlier example described a graph in words and then computed from the description. That is not how the graph questions in an examination work. You are shown a figure, and everything you need — and nothing else — is in it.
These three all read from the same two figures below. Nothing is stated in the text that is not visible in the picture, which is the point: the first skill is extracting the numbers, and it has to be practised on a real drawing.
Read these before reading the questions. Everything needed is here.
On this page
1. Displacement and distance from areas
Problem. From the velocity–time graph above, find the particle’s displacement and the total distance it travels between t = 0 and t = 6 s. Also find its acceleration at t = 1 s and t = 3.5 s, and the instant at which it is momentarily at rest.
The physics
The graph goes negative, so the particle reverses — and the moment it does, displacement and distance stop agreeing. The shading marks the two phases: forward while the curve is above the axis, backward while below. Since the blue region is visibly larger than the red, expect the particle to finish ahead of where it started, but by less than the distance it actually covered.
Before computing, locate the crossing. The line falls from 10 at t = 3 to −5 at t = 4, so it passes through zero somewhere between — and that instant, not t = 4, is where the motion reverses.
The mathematics
Find the crossing. On the segment from (3, 10) to (4, −5) the gradient is −15 m/s per second, so v = 10 − 15(t − 3), which vanishes at
t = 3 + 10/15 = 11/3 ≈ 3.67 s — the instant of rest
Areas above the axis — three pieces, all forward motion:
0 to 2 s, triangle: ½(2)(10) = 10 m
2 to 3 s, rectangle: (1)(10) = 10 m
3 to 11/3 s, triangle: ½(2/3)(10) = 10/3 m
total forward = 70/3 ≈ 23.33 m
Areas below the axis — three pieces, all backward:
11/3 to 4 s, triangle: ½(1/3)(5) = 5/6 m
4 to 5 s, rectangle: (1)(5) = 5 m
5 to 6 s, triangle: ½(1)(5) = 5/2 m
total backward = 25/3 ≈ 8.33 m
Displacement = 70/3 − 25/3 = 15 m
Distance = 70/3 + 25/3 = 95/3 ≈ 31.7 m
Accelerations are gradients. At t = 1 s the particle is on the segment from (0, 0) to (2, 10), so a = 10/2 = +5 m/s². At t = 3.5 s it is on the steep segment from (3, 10) to (4, −5), so a = −15/1 = −15 m/s².
The contemplation
Almost the entire difficulty was finding t = 11/3. The rest is triangles and rectangles that any fourteen-year-old could compute. A student who splits at t = 4 because it is a marked gridline gets 20 m and 30 m — both wrong, both plausible, and nothing in the arithmetic complains. Graph questions punish reading rather than calculating, and the instruction that follows is simple: find where the curve crosses the axis before finding any areas, and do not assume the crossing sits on a gridline.
2. Sketching the position graph from the velocity graph
Problem. Taking the particle to start at x = 0, describe the shape of its position–time graph over the same six seconds: where it rises and falls, where it curves and which way, and its position at each of t = 2, 3, 11/3, 4, 5 and 6 s.
The physics
Two features of the velocity graph carry over, and they carry over to different features of the position graph. The height of the velocity graph becomes the gradient of the position graph — so where v is positive the position rises, and where v is largest it rises most steeply. The gradient of the velocity graph becomes the curvature of the position graph — so where the acceleration is positive the curve bends upward, and where it is zero the position graph is straight.
Both translations matter, and students who remember only the first produce a position graph made entirely of straight lines.
The mathematics
The positions are the running totals of the areas already computed:
x(0) = 0 x(2) = 10 x(3) = 20
x(11/3) = 70/3 ≈ 23.3 x(4) = 22.5
x(5) = 17.5 x(6) = 15 (all in metres)
The shape, section by section.
- 0 to 2 s: rising, and getting steeper — the velocity is increasing from 0 to 10. Curving upward, from a horizontal start.
- 2 to 3 s: rising at a constant 10 m/s. A straight segment, because the acceleration is zero here.
- 3 to 11/3 s: still rising, but flattening rapidly as the velocity falls to zero. Curving downward.
- At t = 11/3 s: momentarily flat — this is the maximum of the position graph, at 23.3 m.
- 11/3 to 4 s: falling, and steepening. Still curving downward.
- 4 to 5 s: falling at a constant 5 m/s. Straight again.
- 5 to 6 s: still falling but flattening, as the velocity returns to zero. Curving upward, ending horizontal at 15 m.
The particle therefore goes out to 23.3 m, turns, and comes back to 15 m — finishing ahead of its start, as the areas already told us.
The contemplation
The maximum of the position graph sits where the velocity graph crosses zero, and nowhere else. That correspondence is worth holding as a single fact rather than two: a turning point on one curve is a zero on the other, because one is the gradient of the other. It also explains why the position graph is smooth where the velocity graph has corners — a corner is a jump in gradient, and the gradient of the position graph is the velocity, which does not jump. Sharp features move up the chain and soften as they go.
3. Equilibrium read off a potential energy curve
Problem. The right-hand figure shows the potential energy of a particle against its position, with three points marked. Classify the equilibrium at A, B and C. State the direction of the force acting on the particle when it sits slightly to the right of A, and slightly to the right of B. Finally, say whether there is any equilibrium at the point of steepest descent between A and B.
The physics
The force is minus the gradient of this curve, so a particle is pushed downhill — toward lower potential energy, always. Everything asked for follows from that one sentence plus the shape of the drawing, and no differentiation is needed anywhere: a flat point is an equilibrium, and whether it is stable depends on whether the ground rises or falls on either side.
The marble picture is exact here. A marble in a bowl returns when nudged; a marble on a dome rolls away; a marble on a level floor stays put wherever you leave it.
The mathematics
- A — a minimum. The curve is flat here, so F = 0 and this is an equilibrium. On both sides the ground rises, so a displaced particle is pushed back. Stable.
- B — a maximum. Flat again, so again an equilibrium. But the ground falls away on both sides, so a displaced particle is pushed further off. Unstable.
- C — a flat region. The curve is level over a whole stretch, not merely at a point, so F = 0 throughout. A displaced particle experiences no force at its new position either, and simply stays there. Neutral.
Just to the right of A, the curve is climbing, so dU/dx > 0 and F = −dU/dx is negative — the force points back toward A, in the −x direction. That is what makes A stable.
Just to the right of B, the curve is descending, so dU/dx < 0 and F is positive — the force points away from B, further to the right. That is what makes B unstable.
The steepest point between A and B. Somewhere on the climb from A to B the curve stops bending one way and starts bending the other. At that point the second derivative is zero — but the first derivative is at its largest, so the force there is the strongest anywhere in the region. It is emphatically not an equilibrium.
This is worth separating carefully from Foundation 8’s warning. There, a zero second derivative at an equilibrium was said to leave the test inconclusive. Here the second derivative is zero at a point that is not an equilibrium at all, because equilibrium is decided by the first derivative and nothing else.
The contemplation
Every answer here came from looking. No function was given, nothing was differentiated, and the classifications are not approximations to a calculation that would have been better — they are the calculation, performed by eye. Foundation 7 claimed that a graph is an instrument rather than an illustration, and this is what that means in practice. Notice also that the same three-way distinction has a shape you already know: dip, hump, plateau. Physics is full of situations where the mathematics is intimidating and the picture is a bowl, a dome and a table.
Reading before computing
All three examples used the same two pictures, and in each the hard part came before the arithmetic: finding the crossing at 11/3 rather than 4, remembering that curvature comes from the velocity’s gradient rather than its height, and recognising that a flat stretch differs from a flat point.
A practical habit for the exam. Before touching a graph question, mark the axis crossings, the turning points and the straight sections on the figure itself. Those three markings contain most of what will be asked, and making them takes about twenty seconds.
Assumes: Foundation 7 · Type: Worked examples · Reading: 18 min · Difficulty: Intermediate