| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a straightforward non-constant acceleration question requiring standard differentiation to find velocity and acceleration, solving a quadratic to find when v=0, then calculating total distance by considering direction changes. All techniques are routine for AS-level mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate \(s\) wrt \(t\) | M1 | AO 3.1a; differentiate with at least 2 powers decreasing by 1 |
| \((v =)\ t^2 - 5t + 6\) | A1 | AO 1.1b; correct expression |
| Equate \(v\) to \(0\) and solve | M1 | AO 1.1b; must attempt to differentiate \(s\) to find \(v\) and solve a 3-term quadratic |
| \(t = 2\) or \(3\) | A1 | AO 1.1b; both values needed |
| \((a =)\ 2t - 5\) | B1ft | AO 2.1; follow their \(v\) (must be differentiating) |
| \(a = 1\) and \(-1\ (\text{m s}^{-2})\) | A1 | AO 1.1b; isw (A0 if extras) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to find values of \(s\) for \(t = 2, 3\) and \(4\) | DM1 | AO 1.1b |
| \(s_2 = \frac{14}{3},\ s_3 = \frac{9}{2},\ s_4 = \frac{16}{3}\) (or differences \(\frac{14}{3},\ -\frac{1}{6},\ \frac{5}{6}\)) | ||
| Total distance \(= s_2 + (s_2 - s_3) + s_4 - s_3\) OR \(s_2 - (s_3 - s_2) + s_4 - s_3\) OR \(\left[\frac{1}{3}t^3 - \frac{5}{2}t^2 + 6t\right]_0^2 - \left[\frac{1}{3}t^3 - \frac{5}{2}t^2 + 6t\right]_2^3 + \left[\frac{1}{3}t^3 - \frac{5}{2}t^2 + 6t\right]_3^4\) OR \(\frac{14}{3} - \left(-\frac{1}{6}\right) + \frac{5}{6}\) | M1 | AO 2.1 |
| \(5\frac{2}{3}\) oe (m) | A1 | AO 1.1b; accept \(5.7\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Clear attempt to find all three \(s\) values (may integrate their \(v\) incorrectly) | DM1 | Dependent on 2nd M1 in part (a); \(t\) values between 0 and 4; no penalty for extra values |
| Complete method using their \(s\) values | M1 | Do NOT condone sign errors |
| \(\frac{17}{3}\) | A1 | Any equivalent fraction, 5.7 or better |
| S.C. \(\int_0^4 \ | t^2 - 5t + 6\ | \, dt = \frac{17}{3}\) |
# Question 3:
## Part 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate $s$ wrt $t$ | M1 | AO 3.1a; differentiate with at least 2 powers decreasing by 1 |
| $(v =)\ t^2 - 5t + 6$ | A1 | AO 1.1b; correct expression |
| Equate $v$ to $0$ and solve | M1 | AO 1.1b; must attempt to differentiate $s$ to find $v$ and solve a 3-term quadratic |
| $t = 2$ or $3$ | A1 | AO 1.1b; both values needed |
| $(a =)\ 2t - 5$ | B1ft | AO 2.1; follow their $v$ (must be differentiating) |
| $a = 1$ and $-1\ (\text{m s}^{-2})$ | A1 | AO 1.1b; isw (A0 if extras) |
## Part 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to find values of $s$ for $t = 2, 3$ and $4$ | DM1 | AO 1.1b |
| $s_2 = \frac{14}{3},\ s_3 = \frac{9}{2},\ s_4 = \frac{16}{3}$ (or differences $\frac{14}{3},\ -\frac{1}{6},\ \frac{5}{6}$) | | |
| Total distance $= s_2 + (s_2 - s_3) + s_4 - s_3$ **OR** $s_2 - (s_3 - s_2) + s_4 - s_3$ **OR** $\left[\frac{1}{3}t^3 - \frac{5}{2}t^2 + 6t\right]_0^2 - \left[\frac{1}{3}t^3 - \frac{5}{2}t^2 + 6t\right]_2^3 + \left[\frac{1}{3}t^3 - \frac{5}{2}t^2 + 6t\right]_3^4$ **OR** $\frac{14}{3} - \left(-\frac{1}{6}\right) + \frac{5}{6}$ | M1 | AO 2.1 |
| $5\frac{2}{3}$ oe (m) | A1 | AO 1.1b; accept $5.7$ or better |
## Question 3b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Clear attempt to find all three $s$ values (may integrate their $v$ incorrectly) | DM1 | Dependent on 2nd M1 in part (a); $t$ values between 0 and 4; no penalty for extra values |
| Complete method using their $s$ values | M1 | Do NOT condone sign errors |
| $\frac{17}{3}$ | A1 | Any equivalent fraction, 5.7 or better |
| **S.C.** $\int_0^4 \|t^2 - 5t + 6\| \, dt = \frac{17}{3}$ | — | Correct answer with no working scores all 3 marks |
---\begin{enumerate}
\item A fixed point $O$ lies on a straight line.
\end{enumerate}
A particle $P$ moves along the straight line.\\
At time $t$ seconds, $t \geqslant 0$, the distance, $s$ metres, of $P$ from $O$ is given by
$$s = \frac { 1 } { 3 } t ^ { 3 } - \frac { 5 } { 2 } t ^ { 2 } + 6 t$$
(a) Find the acceleration of $P$ at each of the times when $P$ is at instantaneous rest.\\
(b) Find the total distance travelled by $P$ in the interval $0 \leqslant t \leqslant 4$
\hfill \mbox{\textit{Edexcel AS Paper 2 2022 Q3 [9]}}