| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a standard variable acceleration question requiring integration of acceleration to find velocity, solving a quadratic for stationary points, and calculating distance with direction changes. While multi-part with several steps, it follows a routine template with straightforward integration, substitution of initial conditions, and area calculation that A-level students practice extensively. Slightly above average due to the need to handle the direction change carefully in part (c). |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \int(4t-9)\,\mathrm{d}t = 2t^2 - 9t\,(+c)\) | M1* | 1.1 - Integrate given expression for \(a\) with at least one term (unsimplified) correct; \(+c\) not required for this first M mark |
| \((1,2) \Rightarrow 2 = 2-9+c \therefore c = \ldots\) | M1dep* | 3.4 - Using given conditions to find \(+c\) |
| \(v = 2t^2 - 9t + 9\) | A1 | 1.1 - Condone '\(v=\)' missing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2t^2 - 9t + 9 = 0 \Rightarrow t = \ldots\) | M1 | 3.4 - Setting their 3-term quadratic for \(v\), from (a), equal to zero and solving for \(t\) |
| \((t_1 =)\ 1.5,\quad (t_2 =)\ 3\) | A1 | 1.1 - BC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{1.5}(2t^2-9t+9)\,\mathrm{d}t = \frac{45}{8}\) | B1FT | 3.1b - BC; correct value for their \(\int_0^{t_1} v\,\mathrm{d}t\). Only FT when their 3-term quadratic in (b) leads to positive values for \(t_1\) and \(t_2\) |
| \(\int_{1.5}^{3}(2t^2-9t+9)\,\mathrm{d}t = -\frac{9}{8}\) | B1FT | 1.1 - BC; correct value for their \(\int_{t_1}^{t_2} v\,\mathrm{d}t\) |
| Total distance travelled is \(6.75\) (m) | B1 | 3.2a - cao |
# Question 10:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \int(4t-9)\,\mathrm{d}t = 2t^2 - 9t\,(+c)$ | M1* | 1.1 - Integrate given expression for $a$ with at least one term (unsimplified) correct; $+c$ not required for this first M mark |
| $(1,2) \Rightarrow 2 = 2-9+c \therefore c = \ldots$ | M1dep* | 3.4 - Using given conditions to find $+c$ |
| $v = 2t^2 - 9t + 9$ | A1 | 1.1 - Condone '$v=$' missing |
**Total: [3]**
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2t^2 - 9t + 9 = 0 \Rightarrow t = \ldots$ | M1 | 3.4 - Setting their 3-term quadratic for $v$, from (a), equal to zero and solving for $t$ |
| $(t_1 =)\ 1.5,\quad (t_2 =)\ 3$ | A1 | 1.1 - BC |
**Total: [2]**
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{1.5}(2t^2-9t+9)\,\mathrm{d}t = \frac{45}{8}$ | B1FT | 3.1b - BC; correct value for their $\int_0^{t_1} v\,\mathrm{d}t$. Only FT when their 3-term quadratic in (b) leads to positive values for $t_1$ and $t_2$ |
| $\int_{1.5}^{3}(2t^2-9t+9)\,\mathrm{d}t = -\frac{9}{8}$ | B1FT | 1.1 - BC; correct value for their $\int_{t_1}^{t_2} v\,\mathrm{d}t$ |
| Total distance travelled is $6.75$ (m) | B1 | 3.2a - cao |
**Total: [3]**
---10 A particle $P$ is moving in a straight line. At time $t$ seconds, where $t \geqslant 0 , P$ has velocity $v \mathrm {~ms} ^ { - 1 }$ and acceleration $a \mathrm {~ms} ^ { - 2 }$ where $a = 4 t - 9$. It is given that $v = 2$ when $t = 1$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$.
The particle $P$ is instantaneously at rest when $t = t _ { 1 }$ and $t = t _ { 2 }$, where $t _ { 1 } < t _ { 2 }$.
\item Find the values of $t _ { 1 }$ and $t _ { 2 }$.
\item Determine the total distance travelled by $P$ between times $t = 0$ and $t = t _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q10 [8]}}