| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward piecewise function question requiring continuity to find k, differentiation for acceleration, and integration for distance. All techniques are standard M2 material with clear signposting and no novel problem-solving required. Slightly easier than average due to explicit guidance and routine calculations. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3}{4}t = \sqrt{2t+1}\) | M1 | Equate the two expressions. Allow M1 only if they verify that it works for \(k=4\) |
| \(9t^2 - 32t - 16 = 0\) | A1 | Correct 3 term quadratic in \(t\) or in \(k\). Any equivalent form without the root |
| \(t = 4\) or \(-\frac{4}{9}\), so \(k = 4\), \(k \geq 0\) | A1* | Given answer for \(k\) correctly explained. The Q asks for an explanation, so they must explain why they reject the negative root |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate \(v\) to obtain \(a\) | M1 | Power decreasing by 1, condone incorrect chain rule |
| \(a = \frac{dv}{dt} = \frac{1}{2}(2t+1)^{-\frac{1}{2}} \times 2\) | A1 | Correct derivative (any equivalent form) |
| When \(t = 1.5\), \(a = 0.5\) (m s\(^{-2}\)) | A1 | cao |
| SC | Allow M1A1A0 for correct differentiation seen as part of a vector approach |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \int \sqrt{2t+1} \, dt\) | M1 | Attempt to integrate: power increasing by 1. Must see working – the question excludes calculators for this step |
| \(= \frac{2}{3}(2t+1)^{\frac{3}{2}} \times \frac{1}{2}\) \((+C)\) | A1 | Correct indefinite integral |
| Correct use of correct limits | M1 | Use of \(t=0\), \(x=0\) and \(t=4\) as limits in a definite integral or to obtain the constant of integration and hence \(x\) when \(t=4\) \(\left(C = -\frac{1}{3}\right)\). "Correct use" means (value when 4 substituted) \(-\) (value when 0 substituted) |
| \(x = \frac{26}{3}\) | A1 | Accept 8.7 or better |
| \(\int \frac{3}{4}t \, dt\) | M1 | Attempt to integrate: power increasing by 1. Must see working. NB It is correct to use \(suvat\) in place of this second interval, but if they do then M1 includes use of the correct initial speed (3 m s\(^{-1}\)) |
| \(= \left[\frac{3}{8}t^2\right]_4^8\) | A1 | Correct definite integral. Accept \(\frac{3 \times 64}{8} - \frac{3 \times 16}{8}\) or equivalent unsimplified expression |
| Total \(= \frac{80}{3}\) (m) \((= 18\) (m)) | A1 | Accept 27 or better \((26.\dot{6})\) |
| SC | Correct integration seen as part of a vector approach can score M1A1M0A0M1A0A0 |
# Question 2:
## Part 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{4}t = \sqrt{2t+1}$ | M1 | Equate the two expressions. Allow M1 only if they verify that it works for $k=4$ |
| $9t^2 - 32t - 16 = 0$ | A1 | Correct 3 term quadratic in $t$ or in $k$. Any equivalent form without the root |
| $t = 4$ or $-\frac{4}{9}$, so $k = 4$, $k \geq 0$ | A1* | **Given answer** for $k$ correctly explained. The Q asks for an explanation, so they must explain why they reject the negative root |
## Part 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate $v$ to obtain $a$ | M1 | Power decreasing by 1, condone incorrect chain rule |
| $a = \frac{dv}{dt} = \frac{1}{2}(2t+1)^{-\frac{1}{2}} \times 2$ | A1 | Correct derivative (any equivalent form) |
| When $t = 1.5$, $a = 0.5$ (m s$^{-2}$) | A1 | cao |
| | SC | Allow M1A1A0 for correct differentiation seen as part of a vector approach |
## Part 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \int \sqrt{2t+1} \, dt$ | M1 | Attempt to integrate: power increasing by 1. Must see working – the question excludes calculators for this step |
| $= \frac{2}{3}(2t+1)^{\frac{3}{2}} \times \frac{1}{2}$ $(+C)$ | A1 | Correct indefinite integral |
| Correct use of correct limits | M1 | Use of $t=0$, $x=0$ and $t=4$ as limits in a definite integral or to obtain the constant of integration and hence $x$ when $t=4$ $\left(C = -\frac{1}{3}\right)$. "Correct use" means (value when 4 substituted) $-$ (value when 0 substituted) |
| $x = \frac{26}{3}$ | A1 | Accept 8.7 or better |
| $\int \frac{3}{4}t \, dt$ | M1 | Attempt to integrate: power increasing by 1. Must see working. NB It is correct to use $suvat$ in place of this second interval, but if they do then M1 includes use of the correct initial speed (3 m s$^{-1}$) |
| $= \left[\frac{3}{8}t^2\right]_4^8$ | A1 | Correct definite integral. Accept $\frac{3 \times 64}{8} - \frac{3 \times 16}{8}$ or equivalent unsimplified expression |
| Total $= \frac{80}{3}$ (m) $(= 18$ (m)) | A1 | Accept 27 or better $(26.\dot{6})$ |
| | SC | Correct integration seen as part of a vector approach can score M1A1M0A0M1A0A0 |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
A particle $P$ is moving in a straight line.\\
At time $t$ seconds, the speed, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of $P$ is given by the continuous function
$$v = \begin{cases} \sqrt { 2 t + 1 } & 0 \leqslant t \leqslant k \\ \frac { 3 } { 4 } t & t > k \end{cases}$$
where $k$ is a constant.\\
(a) Show that $k = 4$, explaining your method carefully.\\
(b) Find the acceleration of $P$ when $t = 1.5$
At time $t = 0 , P$ passes through the point $O$\\
(c) Find the distance of $P$ from $O$ when $t = 8$
\hfill \mbox{\textit{Edexcel M2 2024 Q2 [13]}}