Worked Examples: Drawing Physics

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Module 3: Kinematics  ·  Worked Examples  ·  Prerequisite: K2

Six problems in which the drawing is the answer, and the arithmetic is whatever falls out afterwards.

Mahavakya

A diagram is not a picture of the situation. It is a set of physical claims, and every arrow on it can be wrong.

These are unlike the other worked-example sets. Two ask you to find the false claim in a drawing someone else made. Two ask you to draw first and watch the algebra collapse. Two show the same problem drawn two ways, neither wrong, one far easier.

Work each one with a pencil before reading on. Reading a diagnosis is not the same as making one, and the whole skill here is noticing that something is wrong before being told.

1. What does that arrow claim?

Problem

A block is launched up a smooth slope. A student draws the block with a velocity arrow up the slope and an acceleration arrow also up the slope.

State what that diagram asserts about the motion, and say why it cannot be right.

Two drawings of a block on an incline. On the left, marked wrong, the velocity and acceleration arrows both point up the slope. On the right, marked right, the velocity points up the slope while gravity is drawn vertically downward and the acceleration component along the slope points downhill.

The slope changes what gravity does, not which way gravity points.

What it asserts. Velocity and acceleration pointing the same way means va > 0, which by K4’s rule means the block is speeding up. The diagram claims a block launched uphill accelerates as it climbs.

Why it fails. The only thing acting along the slope is gravity, and gravity is vertical — whatever the slope does. Resolving it gives a component g sin θ directed down the slope, so the acceleration opposes the motion and the block decelerates.

The deeper error is conflating g with a. Some students correct this by drawing gravity along the slope, which is equally wrong for the opposite reason. Gravity’s direction never changes; what the slope changes is how much of it acts along the motion.

Drawing both — g vertical and its component separately — costs one extra arrow and makes the error impossible.

2. What does that shading claim?

Problem

A particle leaves a point at t = 0 and returns to it at t = 6 s. To find the displacement, a student draws a velocity–time graph lying entirely above the axis and shades the area beneath it.

Find the contradiction without computing anything.

Two velocity-time graphs. On the left, marked wrong, the line stays above the axis for the whole interval with the region beneath it shaded. On the right, marked right, the line crosses the axis at three seconds, with the region above shaded green as plus nine metres and the region below shaded red as minus nine metres.

The shading is a claim about the motion, and it can be false.

The contradiction. A curve entirely above the axis says v > 0 throughout — the particle only ever moves forward. But a particle that only moves forward cannot return to where it started.

The drawing contradicts the problem statement, and no arithmetic is needed to see it. The area under that curve is certainly positive, so it cannot be the zero displacement the question describes.

The correct graph crosses the axis. Taking the reversal at t = 3 s:

area above = +9 m    area below = −9 m

displacement = 0    distance = 18 m

Shading is not decoration. Choosing what to shade asserts which regions count positively, and that assertion can be checked against the problem before a single number is written. Here it fails immediately.

3. Two trains, one axis

Problem

Train A moves at 30 m/s and Train B at 20 m/s, heading toward each other on a straight track, initially 1000 m apart.

Draw the system with a single origin and a shared positive direction, then find when they meet.

Two trains on a straight track with a single origin marked at train A and a positive direction arrow pointing right. Train A carries a velocity of plus thirty metres per second and train B, one thousand metres away, carries minus twenty metres per second.

Both signs come from one axis, chosen once.

Put the origin at A’s starting point and take rightward as positive. Then both velocities read off the same axis, and B’s is negative because it moves the other way:

xA = 30t     xB = 1000 − 20t

30t = 1000 − 20t  →  50t = 1000

t = 20 s, at x = 600 m from A’s start

The 50 in the working is the closing speed, and it appeared without being introduced — because the drawing had already fixed the signs. A student who computes 30 + 20 without a diagram is doing the same arithmetic on trust.

The diagram took fifteen seconds and removed every opportunity for a sign error.

4. Brake, then calculate

Problem

A car brakes uniformly from 20 m/s and stops in 40 m.

Mark its position at each whole second before calculating anything, and say what the spacing tells you.

A row of five dots marking a car's position at one-second intervals as it brakes to rest, with the gaps between them measured as seventeen and a half, twelve and a half, seven and a half, and two and a half metres.

The physics is visible in the spacing, before a single equation is written.

The gaps come out 17.5, 12.5, 7.5 and 2.5 m — each 5 m less than the one before.

Shrinking gaps — the car is slowing.

Equal differences between gaps — the deceleration is uniform. Successive differences of 5 m per second per second means |a| = 5 m/s².

Four gaps before the spacing runs out — the car stops after 4 s.

Only now check by equation: v² = u² + 2as gives 0 = 400 − 2a(40), so a = −5 m/s², and t = 20/5 = 4 s. Both confirmed.

The drawing produced the answer before the formula did. Five dots and four measurements gave the sign of the acceleration, its uniformity, its magnitude and the stopping time — none of which required an equation. The equation confirmed what was already visible.

5. Same wedge, two sets of axes

Problem

A block slides on a smooth incline. Draw it twice — once with horizontal and vertical axes, once with axes along and across the slope — and say what each choice costs.

The same block on an incline drawn twice. On the left with horizontal and vertical axes, the acceleration has two components. On the right with axes along and across the slope, the acceleration lies entirely along one axis.

Neither choice is wrong. One of them makes the algebra a single line.

Horizontal and vertical. The acceleration has components in both directions, so you write two equations and eliminate between them. The motion is confined to the slope, which the axes do not know, so that constraint has to be imposed separately.

Along and across the slope. The block cannot leave the surface, so its acceleration across the slope is zero by construction. One equation, no elimination, and the constraint is built into the axes.

Choose the axes that make the constraint simplest, not the ones that make the forces simplest. Here the constraint is “stays on the slope”, and rotating the axes turns it into ay′ = 0 — a statement so simple it disappears.

Both drawings describe the same physics. The difference is entirely in how much work is left afterwards.

6. The same crossing, drawn two ways

Problem

A boat moves at 4 m/s relative to water flowing at 3 m/s. The river is 120 m wide and the boat must land directly opposite its start.

Draw it in the ground frame, then in the water frame, and find the crossing time.

On the left, a river with the boat aiming upstream and travelling straight across to land directly opposite. On the right, the same situation as a right-angled velocity triangle with sides three, four and root seven.

Nothing physical differs between them. The second drawing simply has fewer things in it.

Ground frame. Two velocities to combine, and the boat’s path is not the direction it points — a distinction that has to be held in mind throughout.

Water frame. The current vanishes. The boat moves at 4 m/s, the bank drifts past at 3 m/s, and landing directly opposite means the resultant is perpendicular to the flow. That is a right triangle:

vacross = √(4² − 3²) = √7 m/s

t = 120 / √7 ≈ 45.4 s

The second drawing has three arrows and one right angle. Everything else — the banks, the flow, the two directions the boat might be said to travel in — has been removed, because none of it enters the answer.

Deciding what to leave out is the same skill as deciding what to draw, and it is what makes the second version quicker.

What the six have in common

In examples 1 and 2 the drawing was wrong and the error was findable without arithmetic. In 3 and 4 the drawing produced the answer before the formula did. In 5 and 6 two correct drawings differed only in how much work they left behind.

None of that is illustration. A diagram commits you to an origin, a positive direction, and a claim about every arrow on it — and those commitments are made before any equation exists, which is exactly why they are so easy to make carelessly.

A formula on the page invites scrutiny. A drawing beside it is believed. That asymmetry is the reason this set exists.

Before you calculate

  1. Mark the origin and the positive direction. Both, explicitly.
  2. Draw every vector from the body, labelled with its symbol.
  3. Ask what each arrow claims — and whether you could defend it.
  4. Leave out anything the answer does not need.
  5. When the answer arrives, hold it against the drawing and check they describe the same motion.

Where these sit

These are not graded by difficulty, because the skill is not difficulty-graded — a student who draws well finds all six straightforward and one who does not finds all six hard. Examples 1 and 5 assume K12’s axis choice; example 6 assumes K13; example 4 uses K8’s equations only to confirm what the drawing already showed.

Prerequisite: K2  ·  Working time: 40 min  ·  Level: all

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