| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.3 This is a standard variable acceleration problem requiring integration of a linear acceleration function to find velocity, then solving for rest positions and calculating distance. While it involves multiple steps (integration with initial conditions, solving quadratic equations, and careful distance calculation with direction changes), these are routine M1 techniques with no novel insight required. Slightly easier than average due to straightforward linear acceleration and clear structure. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to integrate \(a\) | M1* | The power of \(t\) must increase by 1 with a change of coefficient in the \(t^2\) term. Do not penalise missing \(c\). Use of \(v = at\) scores M0. |
| \([v =]\ 36t - 3t^2\ [+c]\) or \([v =]\ 36t - \frac{6t^2}{2}\ [+c]\) | A1 | Condone an integral sign in front of correct answer. |
| \(0 = 36t - 3t^2 - 33\); \(\left[27 = 36\times2 - 3\times2^2 + c \Rightarrow c = -33\right]\) | DM1 | Use \(t = 2\) and \(v = 27\) to find \(c\). Must get to \(c =\) and set 3 term quadratic equal to zero. |
| Solve \(0 = 36t - 3t^2 - 33\) to get \(t = 1\) and \(t = 11\) | A1 | Allow \(t = 1\) or \(t = 11\); \(t = 1\), \(t = 11\) oe. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to integrate an expression of the form \(at + bt^2\ [+c]\) with non-zero \(a\) and \(b\). If correct: \([s =]\ \frac{36t^2}{2} - \frac{3t^3}{3} - 33t\ [+c']\) or \([s =]\ 18t^2 - t^3 - 33t\ [+c']\) | M1* | The power of \(t\) must increase by 1 with a change of coefficient in the same term. Use of \(s = vt\) scores M0. |
| Attempt to evaluate their \(\left[18t^2 - t^3 - 33t\right]\) for \(t=0\) to \(t=1\) or \(t=1\) to \(t=11\) or \(t=11\) to \(t=12\); \(0\) to \(1\): \(-16 - 0 = -16\); \(1\) to \(11\): \(484-(-16) = 500\); \(11\) to \(12\): \(468 - 484 = -16\) | DM1 | Attempt using their limits (at least one strictly between 0 and 12) correctly. |
| For all three | DM1 | Allow 11 to 12 implied by symmetry instead of found separately. |
| Distance \(= [16 + 500 + 16] = 532\) m | A1 | — |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to integrate $a$ | M1* | The power of $t$ must increase by 1 with a change of coefficient in the $t^2$ term. Do not penalise missing $c$. Use of $v = at$ scores M0. |
| $[v =]\ 36t - 3t^2\ [+c]$ or $[v =]\ 36t - \frac{6t^2}{2}\ [+c]$ | A1 | Condone an integral sign in front of correct answer. |
| $0 = 36t - 3t^2 - 33$; $\left[27 = 36\times2 - 3\times2^2 + c \Rightarrow c = -33\right]$ | DM1 | Use $t = 2$ and $v = 27$ to find $c$. Must get to $c =$ and set 3 term quadratic equal to zero. |
| Solve $0 = 36t - 3t^2 - 33$ to get $t = 1$ and $t = 11$ | A1 | Allow $t = 1$ or $t = 11$; $t = 1$, $t = 11$ oe. |
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## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to integrate an expression of the form $at + bt^2\ [+c]$ with non-zero $a$ and $b$. If correct: $[s =]\ \frac{36t^2}{2} - \frac{3t^3}{3} - 33t\ [+c']$ or $[s =]\ 18t^2 - t^3 - 33t\ [+c']$ | M1* | The power of $t$ must increase by 1 with a change of coefficient in the same term. Use of $s = vt$ scores M0. |
| Attempt to evaluate their $\left[18t^2 - t^3 - 33t\right]$ for $t=0$ to $t=1$ or $t=1$ to $t=11$ or $t=11$ to $t=12$; $0$ to $1$: $-16 - 0 = -16$; $1$ to $11$: $484-(-16) = 500$; $11$ to $12$: $468 - 484 = -16$ | DM1 | Attempt using their limits (at least one strictly between 0 and 12) correctly. |
| For all three | DM1 | Allow 11 to 12 implied by symmetry instead of found separately. |
| Distance $= [16 + 500 + 16] = 532$ m | A1 | — |6 A particle moves in a straight line. At time $t \mathrm {~s}$, the acceleration, $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, of the particle is given by $a = 36 - 6 t$. The velocity of the particle is $27 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $t = 2$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $t$ when the particle is at instantaneous rest.
\item Find the total distance the particle travels during the first 12 seconds.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q6 [8]}}