1.64PH1 · Section 1: Forces and motion

Stopping distances and terminal velocity

Key learnings

Stopping = thinking + braking distance
Stopping distance = thinking distance + braking distance. Thinking distance = the distance during the driver's reaction time; $d = v \times t_{\text{reaction}}$ ; it rises with speed and reaction time (alcohol, drugs, tiredness). Braking distance = the distance after the brakes are applied; it rises with speed (proportional to $v^2$ ), wet or icy roads, worn tyres and a heavier load. Thinking distance is a driver factor; braking distance is a vehicle/road factor.
Terminal velocity: drag = weight
Terminal velocity = the constant maximum speed of a falling object when drag = weight. Two elements: forces balanced (drag up = weight down) → resultant force = 0 → acceleration = 0, constant velocity. Sequence: at release drag = 0 and weight wins → it accelerates; as speed rises drag grows until drag = weight → terminal velocity. Opening a parachute raises drag above weight, so the diver decelerates to a new, lower terminal velocity.
Reaction time: ruler-drop practical
The ruler-drop practical measures reaction time: one person drops a ruler without warning and the other catches it; the drop distance $d$ gives $t = \sqrt{2d/g}$ (from $d = \tfrac{1}{2}gt^2$ ; $g = 9.8$ m/s², $d$ in m). Typical human reaction time is 0.2–0.4 s. Errors: the catcher anticipating, a signal on release, or parallax when reading the scale — repeat and take the mean to reduce random error. A higher reaction time means a longer thinking distance.

Common mistakes

Drawn from real examiner reports.

Thinking distance rises with speed — say WHY

"Thinking distance increases because the car is going faster" just restates the question. The mark-scheme reason: the reaction time stays the same, but at higher speed the car covers more distance during that fixed time ( $d = v \times t_{\text{reaction}}$ , a linear link). Braking distance, by contrast, rises with the square of speed.

June 2024 Paper 2P examiner report (general guidance): candidates who failed to explain the mechanism behind a change — merely restating the question — scored zero on explanatory marks. This pattern is consistent across all explanation questions in the specification.

Terminal velocity without the force link

"Terminal velocity is when acceleration = 0 / constant speed" scores at most one mark — it gives the consequence, not the force condition. Full form: drag = weight → resultant force = 0 → acceleration = 0 → constant velocity. Worse still is "no force acts": both weight and drag act; they are equal and opposite, giving a zero resultant force (Newton's first law).

November 2023 Paper 2P examiner report Q6(c): principal misconception was that objects require a force to keep moving. At terminal velocity, candidates wrote "no force acts" instead of "forces are balanced" — a fundamental Newton 1 confusion penalised throughout the paper.

v-t graph drawn straight, not a curve

The v-t graph to terminal velocity is a curve: steep at first, then flattening (concave down) as the gradient (acceleration) falls, approaching a horizontal at terminal velocity. A straight line from the origin implies constant acceleration and is wrong. After a parachute opens the velocity curves DOWN to a new, lower plateau — not a vertical drop or a straight line.

June 2023 Paper 2P examiner report (general): graph-drawing questions consistently showed candidates losing marks for drawing straight lines instead of curves — the shape must reflect the changing rate of change, not just the final value.

Braking distance found with d = vt

Braking distance needs a SUVAT equation, not $d = vt$ . The car is decelerating, so use $v^2 = u^2 + 2as$ with $v = 0$ , giving $s = u^2/2a$ . Using $d = vt$ (the thinking-distance formula) is invalid once the brakes are on. Warning sign: if thinking and braking distances come out equal, you have used the wrong formula for one of them.

Ruler drop: cm not converted to m

In $t = \sqrt{2d/g}$ the drop distance $d$ must be in metres. Leaving it in cm gives a factor-of-10 error in the time: an 18 cm drop is $d = 0.18$ m, giving $t \approx 0.19$ s — using 18 gives $t \approx 1.9$ s, which is impossibly long. Convert to metres, then compare against the realistic 0.2–0.4 s range as a check.

Terminal velocity read as zero velocity

At terminal velocity the object is still moving — at a constant (maximum) speed. "Velocity is constant" does NOT mean "velocity is zero". The object has stopped accelerating, not stopped moving. Watch the wording: "it stops accelerating" is right; "it stops" or "it is stationary" is wrong.

Exam tips

Explain via Cause → Mechanism → Effect
For every "explain why" here, use Cause → Mechanism → Effect, one mark per step. E.g. wet road: less friction (cause) → smaller deceleration, $a = F/m$ (mechanism) → larger $s$ from $s = u^2/2a$ (effect). A 3-mark answer needs three distinct linked steps.
Braking distance uses SUVAT, not d = vt
Braking distance comes from $v^2 = u^2 + 2as$ with $v = 0$ (so $s = u^2/2a$ ), while thinking distance uses $d = v \times t_{\text{reaction}}$ . Pick the formula by stage — before braking or after — and never use $d = vt$ once the brakes are on.
Substitute first; show each stage
In multi-step stopping questions, substitute the numbers before rearranging and show each stage separately — the mark scheme awards a mark per correct sub-calculation, so a lone final number risks scoring zero if it is wrong.

Revision notes