| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a standard kinematics question requiring differentiation to find acceleration, solving a quadratic for when v=0, and integration with attention to sign changes. While it requires multiple steps and careful handling of the direction change at t=3, these are routine A-level techniques with no novel insight needed—slightly above average due to the total distance requiring splitting the integral at the stationary point. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Value of \(\dfrac{dv}{dt}\) at \(t=4\) is \(-11\) | B1 | 3.4 |
| Deceleration of \(P\) is \(11\ \text{ms}^{-2}\) | B1 | 2.5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{dv}{dt} = 5 - 4t\) evaluated at \(t=4\) | M1 | |
| Deceleration of \(P\) is \(11\ \text{ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 3\ \text{s}\) | B1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\displaystyle\int_0^3 (2t+1)(3-t)\,dt = \tfrac{27}{2}\) | B1 | 3.1b |
| \(\displaystyle\int_3^4 (2t+1)(3-t)\,dt = -\tfrac{25}{6}\) | B1 | 1.1 |
| Total distance is \(\dfrac{27}{2} + \dfrac{25}{6} = \dfrac{53}{3}\) m or \(17.7\) m | B1 | 1.1 |
## Question 10(a):
Value of $\dfrac{dv}{dt}$ at $t=4$ is $-11$ | **B1** | 3.4 | BC
Deceleration of $P$ is $11\ \text{ms}^{-2}$ | **B1** | 2.5 | Correct positive value, with units
**Alternative solution:**
$\dfrac{dv}{dt} = 5 - 4t$ evaluated at $t=4$ | **M1** |
Deceleration of $P$ is $11\ \text{ms}^{-2}$ | **A1** | | Correct positive value, with units
**[2 marks]**
---
## Question 10(b):
$t = 3\ \text{s}$ | **B1** | 3.4 | $-\frac{1}{2}$, if present, must be rejected
**[1 mark]**
---
## Question 10(c):
$\displaystyle\int_0^3 (2t+1)(3-t)\,dt = \tfrac{27}{2}$ | **B1** | 3.1b | BC
$\displaystyle\int_3^4 (2t+1)(3-t)\,dt = -\tfrac{25}{6}$ | **B1** | 1.1 | BC
Total distance is $\dfrac{27}{2} + \dfrac{25}{6} = \dfrac{53}{3}$ m or $17.7$ m | **B1** | 1.1 | Correct answer (exact, or at least 3sf). If no marks, then award **SC B1** for either correctly integrating $v$ or for an answer of $28/3$ or $9.33$ (at least 3 sf)
**[3 marks]**
---10 A particle $P$ is moving in a straight line. At time $t$ seconds $P$ has velocity $v \mathrm {~ms} ^ { - 1 }$ where $v = ( 2 t + 1 ) ( 3 - t )$.
\begin{enumerate}[label=(\alph*)]
\item Find the deceleration of $P$ when $t = 4$.
\item State the positive value of $t$ for which $P$ is instantaneously at rest.
\item Find the total distance that $P$ travels between times $t = 0$ and $t = 4$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q10 [6]}}