Constant-Acceleration Kinematics and Two-Stage Motion

A parcel slides down a smooth ramp, accelerating uniformly at $2.5$ m s$^{-2}$ from rest for $6$ s, then slides onto a horizontal floor, where friction decelerates it uniformly at $1$ m s$^{-2}$ until it stops. What is the total distance travelled by the parcel?

• $157.5$ m

  Correct. Stage 1: $v = 0+2.5(6)=15$ m/s, $s_1 = \frac{1}{2}(2.5)(6)^2 = 45$ m. Stage 2: from $u=15$ (the velocity carried over from the ramp, not $0$) decelerating at $1$ m/s² to rest takes $t=15$ s, covering $s_2 = \frac{1}{2}(15)(15) = 112.5$ m. Total $=45+112.5=157.5$ m.

• B$45$ m

  This is stage 1's distance alone — the parcel reaching the bottom of the ramp, not the end of its journey. The question describes the floor as part of the same motion, and the parcel keeps travelling (and decelerating) for a further 112.5 m after the ramp.

• C$112.5$ m

  This is stage 2's distance alone, correctly computed using $u=15$ — but the 45 m already covered on the ramp before the parcel ever reached the floor has been left out entirely.

• D$270$ m

  This comes from dropping the $\frac{1}{2}$ in stage 2's area calculation — computing $15 \times 15 = 225$ instead of $\frac{1}{2}(15)(15)=112.5$ — then adding the correct 45 m from stage 1: $225+45=270$. Both stages were correctly identified; the slip is purely inside one line of arithmetic.

Traps tested: Second stage distance omitted · First stage distance omitted · Half dropped from the area formula

A real examiner report on this exact graph shape — acceleration phase, then deceleration phase, over an unknown total time — records: "a common error was to halve the time taken to decelerate rather than double it" (Jan 2023, Q1). Suppose a journey's acceleration phase takes $5$ s, and the deceleration phase is stated to take twice as long as the acceleration phase. What time would a candidate who made exactly this documented error write down for the deceleration phase?

• $2.5$ s — they halved the $5$ s acceleration time instead of doubling it, exactly the error the report names

  Correct. "Twice as long as 5 s" is 10 s; the documented error inverts the operation, producing 5 ÷ 2 = 2.5 s instead. Recognising what a documented error actually produces, not just knowing to avoid it in the abstract, is what makes it catchable in your own working.

• B$10$ s — this is the correct deceleration time, not the value the error produces

  $10$ s is right, but it's the answer to a different question — what SHOULD the deceleration time be, not what does the documented halving error actually produce when applied to these numbers.

• C$20$ s — doubling the stated ratio again, as if "twice as long" meant "four times as long"

  This applies the ratio twice rather than inverting it once. The documented error is specifically a halve-instead-of-double mistake, not a compounding of the given ratio.

• D$5$ s — treating the two phases as equal in length, ignoring the stated ratio entirely

  This discards the "twice as long" relationship altogether rather than inverting it, which is a different (and separately avoidable) mistake from the one the examiner report actually documents.

Traps tested: Correct value mistaken for the error being described · Stated ratio doubled again · Stated ratio ignored entirely

Particle $P$ is thrown vertically upward, its height modelled by $s_P = 20t - 5t^2$, where $t$ is the time in seconds since $P$ was thrown. Particle $Q$ is thrown from the same point $2$ s after $P$, and its height (using its OWN clock, $\tau$ = time since $Q$ was thrown) is modelled by $s_Q = 15\tau - 5\tau^2$. A student wants the value of $t$ (measured on P's clock) at which the two particles are at the same height, and writes $20t - 5t^2 = 15t - 5t^2$, using $t$ in both expressions. What has gone wrong?

• $Q$'s expression needs $\tau = t - 2$ substituted in, not $t$ itself — the two clocks start $2$ s apart and have to be reconciled before the expressions can be equated

  Correct. $s_Q = 15\tau - 5\tau^2$ is a genuinely correct model of $Q$'s height — on $Q$'s own clock. But the question asks for a value of $t$ on $P$'s clock, and at time $t$ on $P$'s clock, $Q$ has only been moving for $t - 2$ seconds. The equation that should be solved is $20t - 5t^2 = 15(t-2) - 5(t-2)^2$, not the one written.

• BNothing — since both expressions model displacement from the same starting point, any shared variable name is valid to equate directly

  This is precisely the documented failure: a real examiner report on this exact question type records candidates who "wrote down two correct expressions for the displacements but failed to realise that they referred to different starting times which led to inconsistent values of t" (Jan 2020, Q3). Each expression can be individually correct and the comparison between them still be invalid.

• CThe error is in $P$'s expression: it should use $\tau = t - 2$ instead, since $Q$ was thrown second

  The adjustment belongs on Q's expression, not P's. $P$'s clock, $t$, is the one the question actually asks for the answer in terms of — it's $Q$'s clock that needs shifting to match it, not the other way round.

• DThe two expressions can never be validly compared, because $P$ and $Q$ have different initial speeds

  Different initial speeds are not the obstacle — plenty of valid two-particle meeting questions involve different speeds. The obstacle is specifically the mismatched starting times, which is fixable (by substituting $\tau = t-2$), not a reason the comparison is impossible outright.

Traps tested: Same time variable reused across different start times · Time shift applied to the wrong particle · Unrelated feature blamed for the real error

A block decelerates uniformly from $10$ m s$^{-1}$ at $2$ m s$^{-2}$ until it comes to rest. While it is moving, its displacement is modelled by $s = 10t - t^2$. Solving $10t - t^2 = 16$ for the time at which the block has travelled $16$ m gives $t = 2$ or $t = 8$. Which is the correct final answer, and why?

• $t = 2$ s only — the block comes to rest at $t = 5$ s (since $v = 10 - 2t = 0$ there), so $t = 8$ falls outside the interval the model $s=10t-t^2$ actually describes, and has to be rejected on that physical ground

  Correct. The equation $s=10t-t^2$ is only a valid description of the block's motion while it is still moving under that deceleration — up to $t=5$s. Beyond that it is at rest, not still obeying the formula, so the mathematical second root at $t=8$ describes a version of the parabola that has no physical counterpart. A real examiner report on this exact structure — a quadratic with one valid and one invalid root — records that "the vast majority found the two roots but some failed to explain clearly why they were rejecting" the invalid one (Oct 2019, Q6): finding both roots is half the task, and stating the reason for discarding one is the other half.

• B$t = 8$ s only, since it is the larger and therefore later of the two times

  Larger is not the same as valid here — $t=8$ s is the one that has to be rejected, not kept, because the block has already been at rest for 3 seconds by the time $t=8$ arrives, and the formula no longer describes anything real by then.

• CBoth $t = 2$ and $t = 8$ — a quadratic equation's roots are both always valid answers to the original question

  A quadratic's two roots are always solutions to the EQUATION — they are not automatically both solutions to the physical QUESTION the equation was built to answer. Here the model itself stops applying at $t=5$s, which silently excludes one of the two mathematically correct roots.

• DNeither — since the equation has two roots, the scenario as described must be impossible

  Two roots from a quadratic is completely ordinary and does not signal an impossible scenario; it signals that one physical constraint (here, the time by which the block has already stopped) needs to be applied to choose between them.

Traps tested: Later root assumed correct without checking validity · Quadratic root rejected without justification · Two roots mistaken for a contradiction

A candidate substitutes correctly into $v = u + at$, but then makes an arithmetic slip and arrives at the wrong final velocity. Based on how M1 mark schemes define M and A marks, what is most likely to happen to their marks for this line?

• They earn the M mark, for correctly applying a suvat equation, but lose the dependent A mark for the wrong numeric answer

  Correct. The real general marking guidance defines the M mark as being for "the application of some mechanical principle to produce an equation... applying a suvat equation," which this candidate genuinely did — the equation itself, and the substitution into it, were both right. The A mark depends on the final number being correct, which it wasn't. This is M1 A0, not M0 A0: the method is still worth something even when the arithmetic that follows it isn't.

• BThey lose the M mark too, since the wrong final answer proves the method itself was flawed

  A wrong final answer doesn't retroactively make a correct method incorrect. The M mark is for the method — the choice of equation and the correct substitution into it — which is exactly what this candidate got right before the slip happened.

• CThey earn both marks, since the method shown was entirely correct

  A marks are described in the general marking guidance as "dependent accuracy... marks," and dependent on the method mark does not mean automatic once the method mark is earned — it still has to be the correct final number. A slipped arithmetic step costs the A mark specifically, not both.

• DWhether the M mark survives depends on how close the wrong answer is to the correct one

  The M mark is earned by the method shown, not graded on how near the final number lands to being right. A method mark is either present (correct method used) or not — proximity of the wrong answer to the right one plays no role in the definition.

Traps tested: Wrong final answer assumed to invalidate a correct method · Method mark mistaken as sufficient for full credit · Mark award conflated with closeness of the wrong answer