| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration from velocity differentiation |
| Difficulty | Moderate -0.8 This is a straightforward M2 question requiring basic differentiation of a quadratic velocity function and substitution. Part (a) is routine calculus (differentiate and substitute), while part (b) requires recognizing the velocity changes sign at t=1/2, making it slightly more involved than pure recall but still a standard textbook exercise with no novel problem-solving required. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(a = \frac{dv}{dt}\) | M1 | Usual rules for differentiation. Condone slip in multiplying brackets |
| \(v = 3t - 2t^2 - 1,\ a = \frac{dv}{dt} = 3 - 4t\) | A1 | |
| \(t = \frac{1}{2},\ a = 1\ (\text{m s}^{-2})\) | A1 | CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = 0 \Rightarrow t = 0.5\) | B1 | Seen or implied |
| \(s = \int 3t - 2t^2 - 1\ dt\) | M1 | Usual rules for integration |
| \(= \frac{3t^2}{2} - \frac{2t^3}{3} - t\ (+C)\ (= F(t))\) | A1ft | Follow their \(v\) |
| Correct strategy for distance | M1 | For their "0.5" in \((0,1)\). Must take account of change in direction |
| \(-[F(t)]_0^{0.5} + [F(t)]_{0.5}^1 = F(1) - 2F(0.5) + F(0)\) | A1 | Or equivalent, accept \(\pm\). For their \(F(t)\) |
| \(\left(= \frac{5}{24} + \frac{1}{24}\right) = 0.25\) m | A1 | CSO |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $a = \frac{dv}{dt}$ | M1 | Usual rules for differentiation. Condone slip in multiplying brackets |
| $v = 3t - 2t^2 - 1,\ a = \frac{dv}{dt} = 3 - 4t$ | A1 | |
| $t = \frac{1}{2},\ a = 1\ (\text{m s}^{-2})$ | A1 | CSO |
---
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = 0 \Rightarrow t = 0.5$ | B1 | Seen or implied |
| $s = \int 3t - 2t^2 - 1\ dt$ | M1 | Usual rules for integration |
| $= \frac{3t^2}{2} - \frac{2t^3}{3} - t\ (+C)\ (= F(t))$ | A1ft | Follow their $v$ |
| Correct strategy for distance | M1 | For their "0.5" in $(0,1)$. Must take account of change in direction |
| $-[F(t)]_0^{0.5} + [F(t)]_{0.5}^1 = F(1) - 2F(0.5) + F(0)$ | A1 | Or equivalent, accept $\pm$. For their $F(t)$ |
| $\left(= \frac{5}{24} + \frac{1}{24}\right) = 0.25$ m | A1 | CSO |
---2. A particle $P$ moves in a straight line. At time $t = 0 , P$ passes through a point $O$ on the line. At time $t$ seconds, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ where
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ when $t = \frac { 1 } { 2 }$
\item Find the distance travelled by $P$ in the interval $0 \leqslant t \leqslant 1$
At time $t$ seconds, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ where
$$v = ( 2 t - 1 ) ( 1 - t )$$
(a) Find the acceleration of $P$ when $t = \frac { 1 } { 2 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q2 [9]}}