| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find position by integrating velocity |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring standard integration and differentiation of vector functions with simple powers. Part (a) involves substitution and magnitude calculation, (b) is direct differentiation, and (c) requires integration with a constant of integration found from given conditions. All techniques are routine A-level mechanics procedures with no conceptual challenges or novel problem-solving required. |
| 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Put \(t = 2\) in v and use Pythagoras: \(\sqrt{12^2 + (-6\sqrt{2})^2}\) | M1 | Need square root but negative sign not required. Allow i's and/or j's to go missing from their v at \(t = 2\), provided they have applied Pythagoras correctly. |
| \(\sqrt{216}, 6\sqrt{6}\) or \(15\) or better \((\text{m s}^{-1})\) | A1 | cao. Correct answer with no working can score 2 marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate v wrt \(t\) to obtain a | M1 | Both powers decreasing by 1. Allow a column vector. M0 if i or j is missing but allow recovery in (b). |
| \(6t\mathbf{i} - 3t^{-\frac{1}{2}}\mathbf{j}\) oe \((\text{m s}^{-2})\) isw | A1 | cao. Do not accept a column vector. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrate v wrt \(t\) to obtain r | M1 | Both powers increasing by 1. M0 if i or j is missing but allow recovery. |
| \(\mathbf{r} = t^3\mathbf{i} - 4t^{\frac{3}{2}}\mathbf{j}\ (+\mathbf{C})\) | A1 | \((\mathbf{r} =)\) not required |
| \((\mathbf{i} - 4\mathbf{j}) = 4^3\mathbf{i} - 4 \times 4^{\frac{3}{2}}\mathbf{j} + \mathbf{C}\) | M1 | Putting \(\mathbf{r} = (\mathbf{i} - 4\mathbf{j})\) and \(t = 4\) into displacement vector expression which must have C (allow \(C\)) to give an equation in C only, seen or implied. Must have attempted to integrate v. N.B. C does not need to be found — this is a method mark, so allow slips. |
| \((-62\mathbf{i} + 24\mathbf{j})\) (m) isw | A1 | cao |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Put $t = 2$ in **v** and use Pythagoras: $\sqrt{12^2 + (-6\sqrt{2})^2}$ | M1 | Need square root but negative sign not required. Allow **i**'s and/or **j**'s to go missing from their **v** at $t = 2$, provided they have applied Pythagoras correctly. |
| $\sqrt{216}, 6\sqrt{6}$ or $15$ or better $(\text{m s}^{-1})$ | A1 | cao. Correct answer with no working can score 2 marks. |
**(2 marks)**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate **v** wrt $t$ to obtain **a** | M1 | Both powers decreasing by 1. Allow a column vector. M0 if **i** or **j** is missing but allow recovery in (b). |
| $6t\mathbf{i} - 3t^{-\frac{1}{2}}\mathbf{j}$ oe $(\text{m s}^{-2})$ isw | A1 | cao. Do not accept a column vector. |
**(2 marks)**
---
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate **v** wrt $t$ to obtain **r** | M1 | Both powers increasing by 1. M0 if **i** or **j** is missing but allow recovery. |
| $\mathbf{r} = t^3\mathbf{i} - 4t^{\frac{3}{2}}\mathbf{j}\ (+\mathbf{C})$ | A1 | $(\mathbf{r} =)$ not required |
| $(\mathbf{i} - 4\mathbf{j}) = 4^3\mathbf{i} - 4 \times 4^{\frac{3}{2}}\mathbf{j} + \mathbf{C}$ | M1 | Putting $\mathbf{r} = (\mathbf{i} - 4\mathbf{j})$ and $t = 4$ into displacement **vector** expression which must have **C** (allow $C$) to give an equation in **C** only, seen or implied. Must have attempted to integrate **v**. **N.B.** **C** does not need to be found — this is a method mark, so allow slips. |
| $(-62\mathbf{i} + 24\mathbf{j})$ (m) isw | A1 | cao |
**(4 marks)**
**(8 marks total)**\begin{enumerate}
\item \hspace{0pt} [In this question, position vectors are given relative to a fixed origin.] At time $t$ seconds, where $t > 0$, a particle $P$ has velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ where
\end{enumerate}
$$\mathbf { v } = 3 t ^ { 2 } \mathbf { i } - 6 t ^ { \frac { 1 } { 2 } } \mathbf { j }$$
(a) Find the speed of $P$ at time $t = 2$ seconds.\\
(b) Find an expression, in terms of $t , \mathbf { i }$ and $\mathbf { j }$, for the acceleration of $P$ at time $t$ seconds, where $t > 0$
At time $t = 4$ seconds, the position vector of $P$ is ( $\mathbf { i } - 4 \mathbf { j }$ ) m.\\
(c) Find the position vector of $P$ at time $t = 1$ second.
\hfill \mbox{\textit{Edexcel Paper 3 2022 Q1 [8]}}