| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find position by integrating velocity |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring standard integration and differentiation of vector functions with fractional powers. Part (a) involves differentiating velocity to find acceleration, and part (b) requires integrating velocity and applying initial conditions. Both are routine A-level Further Maths mechanics techniques with no problem-solving insight needed, making it easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Differentiate v | M1 | Use of \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\) with attempt to differentiate (both powers decreasing by 1). M0 if i's and j's omitted and they don't recover |
| \(\mathbf{a} = 6\mathbf{i} - \frac{15}{2}t^{\frac{1}{2}}\mathbf{j}\) | A1 | Correct differentiation in any form |
| \(= 6\mathbf{i} - 15\mathbf{j}\) (m s\(^{-2}\)) | A1 | Correct and simplified. ISW if they go on and find the magnitude |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Integrate v | M1 | Use of \(\mathbf{r} = \int \mathbf{v}\, dt\) with attempt to integrate (both powers increasing by 1). M0 if i's and j's omitted and they don't recover |
| \(\mathbf{r} = (\mathbf{r}_0) + 3t^2\mathbf{i} - 2t^{\frac{5}{2}}\mathbf{j}\) | A1 | Correct integration in any form. Condone \(\mathbf{r}_0\) not present |
| \(= (-20\mathbf{i} + 20\mathbf{j}) + (48\mathbf{i} - 64\mathbf{j}) = 28\mathbf{i} - 44\mathbf{j}\) (m) | A1 | Correct and simplified |
| (3) | ||
| Total: (6) | N.B. Accept column vectors throughout and condone missing brackets in working but they must be there in final answers |
## Question 1:
### Part 1(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Differentiate **v** | M1 | Use of $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ with attempt to differentiate (both powers decreasing by 1). M0 if **i**'s and **j**'s omitted and they don't recover |
| $\mathbf{a} = 6\mathbf{i} - \frac{15}{2}t^{\frac{1}{2}}\mathbf{j}$ | A1 | Correct differentiation in any form |
| $= 6\mathbf{i} - 15\mathbf{j}$ (m s$^{-2}$) | A1 | Correct and simplified. ISW if they go on and find the magnitude |
| **(3)** | | |
### Part 1(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Integrate **v** | M1 | Use of $\mathbf{r} = \int \mathbf{v}\, dt$ with attempt to integrate (both powers increasing by 1). M0 if **i**'s and **j**'s omitted and they don't recover |
| $\mathbf{r} = (\mathbf{r}_0) + 3t^2\mathbf{i} - 2t^{\frac{5}{2}}\mathbf{j}$ | A1 | Correct integration in any form. Condone $\mathbf{r}_0$ not present |
| $= (-20\mathbf{i} + 20\mathbf{j}) + (48\mathbf{i} - 64\mathbf{j}) = 28\mathbf{i} - 44\mathbf{j}$ (m) | A1 | Correct and simplified |
| **(3)** | | |
| **Total: (6)** | | N.B. Accept column vectors throughout and condone missing brackets in working but they must be there in final answers |\begin{enumerate}
\item \hspace{0pt} [In this question position vectors are given relative to a fixed origin $O$ ]
\end{enumerate}
At time $t$ seconds, where $t \geqslant 0$, a particle, $P$, moves so that its velocity $\mathbf { v ~ m ~ s } ^ { - 1 }$ is given by
$$\mathbf { v } = 6 t \mathbf { i } - 5 t ^ { \frac { 3 } { 2 } } \mathbf { j }$$
When $t = 0$, the position vector of $P$ is $( - 20 \mathbf { i } + 20 \mathbf { j } ) \mathrm { m }$.\\
(a) Find the acceleration of $P$ when $t = 4$\\
(b) Find the position vector of $P$ when $t = 4$\\
\hfill \mbox{\textit{Edexcel Paper 3 2019 Q1 [6]}}