| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position from velocity and initial conditions |
| Difficulty | Moderate -0.3 This is a straightforward kinematics question using constant acceleration equations (suvat) in vector form. Part (a) is a 'show that' requiring one calculation (v = u + at), parts (b-d) apply standard formulas with clear initial conditions. While it has multiple parts and requires careful vector component work, it involves only routine application of well-practiced techniques with no conceptual challenges or novel problem-solving. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\): \((10.5\mathbf{i} - 0.9\mathbf{j}) = 0.6\mathbf{j} + 15\mathbf{a}\) | M1 | For use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) |
| \(\mathbf{a} = (0.7\mathbf{i} - 0.1\mathbf{j})\) m s\(^{-2}\) Given answer | A1 | For given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) | M1 | For use of \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) |
| \(\mathbf{r} = 0.6\mathbf{j}\, t + \dfrac{1}{2}(0.7\mathbf{i} - 0.1\mathbf{j})\,t^2\) | A1 | For a correct expression for \(\mathbf{r}\) in terms of \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Equating the \(\mathbf{i}\) and \(\mathbf{j}\) components of \(\mathbf{r}\) | M1 | For equating the \(\mathbf{i}\) and \(\mathbf{j}\) components of their \(\mathbf{r}\) |
| \(\dfrac{1}{2}(0.7)\,t^2 = 0.6\,t - \dfrac{1}{2}(0.1)\,t^2\) | A1ft | For a correct equation following their \(\mathbf{r}\) |
| \(t = 1.5\) | A1 | For \(t = 1.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\): \(\mathbf{v} = 0.6\mathbf{j} + (0.7\mathbf{i} - 0.1\mathbf{j})\,t\) | M1 | For use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) for a general \(t\) |
| Equating the \(\mathbf{i}\) and \(\mathbf{j}\) components of \(\mathbf{v}\) | M1 | For equating the \(\mathbf{i}\) and \(\mathbf{j}\) components of their \(\mathbf{v}\) |
| \(t = 0.75\) | A1ft | For \(t = 0.75\), or correct follow through from incorrect equation |
## Question 8:
### Part (a):
| Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$: $(10.5\mathbf{i} - 0.9\mathbf{j}) = 0.6\mathbf{j} + 15\mathbf{a}$ | M1 | For use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ |
| $\mathbf{a} = (0.7\mathbf{i} - 0.1\mathbf{j})$ m s$^{-2}$ Given answer | A1 | For given answer correctly obtained |
### Part (b):
| Use of $\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2$ | M1 | For use of $\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2$ |
| $\mathbf{r} = 0.6\mathbf{j}\, t + \dfrac{1}{2}(0.7\mathbf{i} - 0.1\mathbf{j})\,t^2$ | A1 | For a correct expression for $\mathbf{r}$ in terms of $t$ |
### Part (c):
| Equating the $\mathbf{i}$ and $\mathbf{j}$ components of $\mathbf{r}$ | M1 | For equating the $\mathbf{i}$ and $\mathbf{j}$ components of their $\mathbf{r}$ |
| $\dfrac{1}{2}(0.7)\,t^2 = 0.6\,t - \dfrac{1}{2}(0.1)\,t^2$ | A1ft | For a correct equation following their $\mathbf{r}$ |
| $t = 1.5$ | A1 | For $t = 1.5$ |
### Part (d):
| Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$: $\mathbf{v} = 0.6\mathbf{j} + (0.7\mathbf{i} - 0.1\mathbf{j})\,t$ | M1 | For use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ for a general $t$ |
| Equating the $\mathbf{i}$ and $\mathbf{j}$ components of $\mathbf{v}$ | M1 | For equating the $\mathbf{i}$ and $\mathbf{j}$ components of their $\mathbf{v}$ |
| $t = 0.75$ | A1ft | For $t = 0.75$, or correct follow through from incorrect equation |
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors due east and due north respectively]
\end{enumerate}
A radio controlled model boat is placed on the surface of a large pond.\\
The boat is modelled as a particle.\\
At time $t = 0$, the boat is at the fixed point $O$ and is moving due north with speed $0.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
Relative to $O$, the position vector of the boat at time $t$ seconds is $\mathbf { r }$ metres.\\
At time $t = 15$, the velocity of the boat is $( 10.5 \mathbf { i } - 0.9 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
The acceleration of the boat is constant.\\
(a) Show that the acceleration of the boat is $( 0.7 \mathbf { i } - 0.1 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$.\\
(b) Find $\mathbf { r }$ in terms of $t$.\\
(c) Find the value of $t$ when the boat is north-east of $O$.\\
(d) Find the value of $t$ when the boat is moving in a north-east direction.
\hfill \mbox{\textit{Edexcel Paper 3 Q8 [10]}}