| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.3 This is a straightforward kinematics question requiring differentiation of velocity to find acceleration and integration to find distance. Part (a) involves setting v=0 to find when at rest, then differentiating v to get acceleration. Part (b) requires checking if the particle changes direction and integrating appropriately. These are standard AS-level mechanics techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(v = 12 + 4t - t^2 = 0\) and solving | M1 | Equating \(v\) to 0 and solving the quadratic. If no evidence of solving and at least one answer wrong, M0 |
| \(t = 6\) (or \(-2\)) | A1 | 6 but allow \(-2\) as well at this stage |
| Differentiate \(v\) wrt \(t\) | M1 | For differentiation (both powers decreasing by 1) |
| \(\left(a = \frac{dv}{dt} =\right) 4 - 2t\) | A1 | cao; only need RHS |
| When \(t = 6\), \(a = -8\); Magnitude is \(8\) (m s\(^{-2}\)) | A1 | Substitute \(t=6\) and get 8 as answer. Must be positive. (A0 if two answers given) |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Integrate \(v\) wrt \(t\) | M1 | For integration (at least two powers increasing by 1) |
| \(\left(s =\right) 12t + 2t^2 - \frac{1}{3}t^3 (+C)\) | A1 | Correct expression (ignore \(C\)), only need RHS. Must be used in part (b) |
| \(t = 3 \Rightarrow\) distance \(= 45\) (m) | A1 | Correct distance. Ignore units |
| (3) | ||
| (8 marks) |
## Question 3:
**Part (a):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $v = 12 + 4t - t^2 = 0$ and solving | M1 | Equating $v$ to 0 and solving the quadratic. If no evidence of solving and at least one answer wrong, M0 |
| $t = 6$ (or $-2$) | A1 | 6 but allow $-2$ as well at this stage |
| Differentiate $v$ wrt $t$ | M1 | For differentiation (both powers decreasing by 1) |
| $\left(a = \frac{dv}{dt} =\right) 4 - 2t$ | A1 | cao; only need RHS |
| When $t = 6$, $a = -8$; Magnitude is $8$ (m s$^{-2}$) | A1 | Substitute $t=6$ and get 8 as answer. Must be **positive**. (A0 if two answers given) |
| **(5)** | | |
**Part (b):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Integrate $v$ wrt $t$ | M1 | For integration (at least two powers increasing by 1) |
| $\left(s =\right) 12t + 2t^2 - \frac{1}{3}t^3 (+C)$ | A1 | Correct expression (ignore $C$), only need RHS. Must be used in part (b) |
| $t = 3 \Rightarrow$ distance $= 45$ (m) | A1 | Correct distance. Ignore units |
| **(3)** | | |
| **(8 marks)** | | |\begin{enumerate}
\item A particle, $P$, moves along a straight line such that at time $t$ seconds, $t \geqslant 0$, the velocity of $P$, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is modelled as
\end{enumerate}
$$v = 12 + 4 t - t ^ { 2 }$$
Find\\
(a) the magnitude of the acceleration of $P$ when $P$ is at instantaneous rest,\\
(b) the distance travelled by $P$ in the interval $0 \leqslant t \leqslant 3$
\hfill \mbox{\textit{Edexcel AS Paper 2 2019 Q3 [8]}}