| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a standard variable acceleration question requiring integration of a^(1/2) with given boundary conditions, followed by routine distance calculations and finding maximum acceleration. All steps are mechanical applications of calculus techniques commonly practiced in M1, with no novel problem-solving required beyond following the standard procedure of integrating acceleration to get velocity and velocity to get displacement. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For attempt at integration | M1 | The power of \(t\) must increase by 1 with a change of coefficient. Do not penalise missing \(c\). Use of \(v = at\) scores M0. |
| \(v = \frac{2}{3}kt^{\frac{3}{2}}[+c]\) | A1 | Allow unsimplified e.g. \(v = \frac{1}{1.5}kt^{\frac{1}{2}+1}[+c]\). |
| \(1.8 = \frac{2}{3}k\times 9^{\frac{3}{2}} \Rightarrow k = \left[\frac{3}{2}\times 1.8 \div 27 =\right] 0.1\) | B1 | AG. Must show values substituted OE (e.g. \(1.8 = 18k\)). |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For attempt at integration of either \(\int\left(\frac{2}{3}kt^{\frac{3}{2}}\right)dt\) or \(\int\left(0.2(t-9)^2+1.8\right)dt\) or \(\int\left(0.2t^2-3.6t+18\right)dt\) | M1 | The power of \(t\) or \((t-9)\) must increase by 1 with a change of coefficient in at least one term. Use of \(s=vt\) is M0. |
| \(\left[\frac{4}{150}t^{\frac{5}{2}}\right]_0^9\) and \(\left[\left(\frac{0.2}{3}(t-9)^3+1.8t\right)\right]_9^{18}\) or \(\left[\left(\frac{0.2}{3}t^3-\frac{3.6}{2}t^2+18t\right)\right]_9^{18}\) | A1 | Allow unsimplified. No need for limits. Could include \(+c\) with either or both. |
| \(=\frac{4}{150}\times 9^{\frac{5}{2}}\left[=6.48\text{ or }\frac{162}{25}\right]\) or \(=\frac{0.2}{3}(18-9)^3+1.8\times18-1.8\times9=\left[64.8\text{ or }\frac{324}{5}\right]\) or \(\left(\frac{0.2}{3}\times18^3-\frac{3.6}{2}\times18^2+18\times18\right)-\left(\frac{0.2}{3}\times9^3-\frac{3.6}{2}\times9^2+18\times9\right)\) \(\left[=\frac{648}{5}-\frac{324}{5}=\frac{324}{5}\right]\) | M1 | Correct use of limits 0, 9 or limits 9, 18. Can be implied by either answer following integration. \(+c\) method: \(s=\frac{0.2}{3}(t-9)^3+1.8t-\frac{243}{25}\), or \(s=\frac{0.2}{3}t^3-\frac{3.6}{2}t^2+18t-\frac{1458}{25}\), then substitute \(t=18\) to get \(\frac{1782}{25}\), subtract \(\frac{162}{25}\), so distance \(=\frac{324}{5}\). |
| \(6.48=\frac{1}{10}\times64.8\) or \(64.8=10\times6.48\) or \(\frac{324}{5}=10\times\frac{162}{25}\) | A1 | AG OE. Check working as can get answer from wrong working. NFWW. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either \(\int_0^9\left(\frac{2}{3}kt^{\frac{3}{2}}\right)dt=6.48\) or \(\frac{162}{25}\); Or \(\int_9^{18}\left(0.2(t-9)^2+1.8\right)dt=64.8\) or \(\frac{324}{5}\) | SC B1 | |
| \(6.48=\frac{1}{10}\times64.8\) or \(64.8=10\times6.48\) or \(\frac{324}{5}=10\times\frac{162}{25}\) | SC B1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For differentiation; should get \(a=0.4(t-9)\) or \(a=0.4t-3.6\) | M1 | The power of \(t\) or \((t-9)\) must decrease by 1 with a change of coefficient. M0 for \(a=\frac{v}{t}\). |
| \(0.4\ [\text{ms}^{-2}]\) [at \(t=10\)] | A1 | SC B1 for 0.4 with no differentiation seen. |
| 0.3 seen (from first phase) and state that 0.4 is final answer | B1 | No working needed. If M1A0 or M0A0 scored, then SC B1 for 0.3 without mention of the maximum acceleration. |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| For attempt at integration | M1 | The power of $t$ must increase by 1 with a change of coefficient. Do not penalise missing $c$. Use of $v = at$ scores M0. |
| $v = \frac{2}{3}kt^{\frac{3}{2}}[+c]$ | A1 | Allow unsimplified e.g. $v = \frac{1}{1.5}kt^{\frac{1}{2}+1}[+c]$. |
| $1.8 = \frac{2}{3}k\times 9^{\frac{3}{2}} \Rightarrow k = \left[\frac{3}{2}\times 1.8 \div 27 =\right] 0.1$ | B1 | AG. Must show values substituted OE (e.g. $1.8 = 18k$). |
| | **3** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For attempt at integration of either $\int\left(\frac{2}{3}kt^{\frac{3}{2}}\right)dt$ or $\int\left(0.2(t-9)^2+1.8\right)dt$ or $\int\left(0.2t^2-3.6t+18\right)dt$ | **M1** | The power of $t$ or $(t-9)$ must increase by 1 with a change of coefficient in at least one term. Use of $s=vt$ is M0. |
| $\left[\frac{4}{150}t^{\frac{5}{2}}\right]_0^9$ and $\left[\left(\frac{0.2}{3}(t-9)^3+1.8t\right)\right]_9^{18}$ or $\left[\left(\frac{0.2}{3}t^3-\frac{3.6}{2}t^2+18t\right)\right]_9^{18}$ | **A1** | Allow unsimplified. No need for limits. Could include $+c$ with either or both. |
| $=\frac{4}{150}\times 9^{\frac{5}{2}}\left[=6.48\text{ or }\frac{162}{25}\right]$ or $=\frac{0.2}{3}(18-9)^3+1.8\times18-1.8\times9=\left[64.8\text{ or }\frac{324}{5}\right]$ or $\left(\frac{0.2}{3}\times18^3-\frac{3.6}{2}\times18^2+18\times18\right)-\left(\frac{0.2}{3}\times9^3-\frac{3.6}{2}\times9^2+18\times9\right)$ $\left[=\frac{648}{5}-\frac{324}{5}=\frac{324}{5}\right]$ | **M1** | Correct use of limits 0, 9 or limits 9, 18. Can be implied by either answer following integration. $+c$ method: $s=\frac{0.2}{3}(t-9)^3+1.8t-\frac{243}{25}$, or $s=\frac{0.2}{3}t^3-\frac{3.6}{2}t^2+18t-\frac{1458}{25}$, then substitute $t=18$ to get $\frac{1782}{25}$, subtract $\frac{162}{25}$, so distance $=\frac{324}{5}$. |
| $6.48=\frac{1}{10}\times64.8$ or $64.8=10\times6.48$ or $\frac{324}{5}=10\times\frac{162}{25}$ | **A1** | AG OE. Check working as can get answer from wrong working. NFWW. |
**Special Case (calculator use): Award max 2/4**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Either $\int_0^9\left(\frac{2}{3}kt^{\frac{3}{2}}\right)dt=6.48$ or $\frac{162}{25}$; Or $\int_9^{18}\left(0.2(t-9)^2+1.8\right)dt=64.8$ or $\frac{324}{5}$ | **SC B1** | |
| $6.48=\frac{1}{10}\times64.8$ or $64.8=10\times6.48$ or $\frac{324}{5}=10\times\frac{162}{25}$ | **SC B1** | OE |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For differentiation; should get $a=0.4(t-9)$ or $a=0.4t-3.6$ | **M1** | The power of $t$ or $(t-9)$ must decrease by 1 with a change of coefficient. M0 for $a=\frac{v}{t}$. |
| $0.4\ [\text{ms}^{-2}]$ [at $t=10$] | **A1** | SC B1 for 0.4 with no differentiation seen. |
| 0.3 seen (from first phase) and state that 0.4 is final answer | **B1** | No working needed. If M1A0 or M0A0 scored, then SC B1 for 0.3 without mention of the maximum acceleration. |
---5 A particle starts from rest from a point $O$ and moves in a straight line. The acceleration of the particle at time $t \mathrm {~s}$ after leaving $O$ is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $a = k t ^ { \frac { 1 } { 2 } }$ for $0 \leqslant t \leqslant 9$ and where $k$ is a constant. The velocity of the particle at $t = 9$ is $1.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = 0.1$.\\
For $t > 9$, the velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of the particle is given by $v = 0.2 ( t - 9 ) ^ { 2 } + 1.8$.
\item Show that the distance travelled in the first 9 seconds is one tenth of the distance travelled between $t = 9$ and $t = 18$.
\item Find the greatest acceleration of the particle during the first 10 seconds of its motion.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q5 [10]}}