| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Velocity from acceleration and initial conditions |
| Difficulty | Moderate -0.3 This is a standard two-part kinematics question using constant acceleration equations (suvat) in vector form. Part (a) requires finding acceleration from initial velocity and displacement (routine calculation), while part (b) involves finding time when velocity components satisfy a directional constraint (i=j for northeast). Both parts use well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\): \((7\mathbf{i} - 10\mathbf{j}) = 2(2\mathbf{i} - 3\mathbf{j}) + \frac{1}{2}\mathbf{a}2^2\) | M1 | 3.1b |
| \(\mathbf{a} = (1.5\mathbf{i} - 2\mathbf{j})\) | A1 | 1.1b |
| \( | \mathbf{a} | = \sqrt{1.5^2 + (-2)^2}\) |
| \(= 2.5\) m s\(^{-2}\) * GIVEN ANSWER | A1* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t = (2\mathbf{i}-3\mathbf{j}) + 2(1.5\mathbf{i}-2\mathbf{j})\) | M1 | 3.1b |
| \(= (5\mathbf{i} - 7\mathbf{j})\) | A1 | 1.1b |
| \(\mathbf{v} = (5\mathbf{i}-7\mathbf{j}) + t(4\mathbf{i}+8.8\mathbf{j}) = (5+4t)\mathbf{i} + (8.8t-7)\mathbf{j}\) and \((5+4t) = (8.8t-7)\) | M1 | 3.1b |
| \(t = 2.5\) (s) | A1 | 1.1b |
## Question 8:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$: $(7\mathbf{i} - 10\mathbf{j}) = 2(2\mathbf{i} - 3\mathbf{j}) + \frac{1}{2}\mathbf{a}2^2$ | M1 | 3.1b |
| $\mathbf{a} = (1.5\mathbf{i} - 2\mathbf{j})$ | A1 | 1.1b |
| $|\mathbf{a}| = \sqrt{1.5^2 + (-2)^2}$ | M1 | 1.1b |
| $= 2.5$ m s$^{-2}$ * GIVEN ANSWER | A1* | 2.1 |
**Notes:**
- M1: Using a complete method to obtain the acceleration. Equation in **a** only, could be obtained by two integrations
- Alternative: Use velocity at half-time ($t=1$) = Average velocity over time period. So at $t=1$, $\mathbf{v} = \frac{1}{2}(7\mathbf{i}-10\mathbf{j})$ so $\mathbf{a} = \frac{1}{2}(7\mathbf{i}-10\mathbf{j})-(2\mathbf{i}-3\mathbf{j})$
- A1: Correct **a** vector
- M1: Attempt to find magnitude using $\sqrt{a^2+b^2}$
- A1*: Correct GIVEN ANSWER obtained correctly
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t = (2\mathbf{i}-3\mathbf{j}) + 2(1.5\mathbf{i}-2\mathbf{j})$ | M1 | 3.1b |
| $= (5\mathbf{i} - 7\mathbf{j})$ | A1 | 1.1b |
| $\mathbf{v} = (5\mathbf{i}-7\mathbf{j}) + t(4\mathbf{i}+8.8\mathbf{j}) = (5+4t)\mathbf{i} + (8.8t-7)\mathbf{j}$ and $(5+4t) = (8.8t-7)$ | M1 | 3.1b |
| $t = 2.5$ (s) | A1 | 1.1b |
**Notes:**
- M1: Using a complete method to obtain velocity at $A$, e.g. use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ with $t=2$ and $\mathbf{u} = 2\mathbf{i}-3\mathbf{j}$ and their **a**
- OR: by use of $\mathbf{s} = \mathbf{v}t - \frac{1}{2}\mathbf{a}t^2$
- OR: by integrating their **a**, with addition of $\mathbf{C} = 2\mathbf{i}-3\mathbf{j}$, and putting $t=2$
- A1: correct vector
- M1: Complete method to find equation in $t$ only, e.g. by using $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ and equating **i** and **j** components. Must be equating **i** and **j** components of a velocity vector and must be their velocity at $A$
- A1: 2.5 (s)
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point $O$.]
\end{enumerate}
A particle $P$ moves with constant acceleration.\\
At time $t = 0$, the particle is at $O$ and is moving with velocity ( $2 \mathbf { i } - 3 \mathbf { j }$ ) $\mathrm { ms } ^ { - 1 }$\\
At time $t = 2$ seconds, $P$ is at the point $A$ with position vector ( $7 \mathbf { i } - 10 \mathbf { j }$ ) m.\\
(a) Show that the magnitude of the acceleration of $P$ is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
At the instant when $P$ leaves the point $A$, the acceleration of $P$ changes so that $P$ now moves with constant acceleration ( $4 \mathbf { i } + 8.8 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 2 }$
At the instant when $P$ reaches the point $B$, the direction of motion of $P$ is north east.\\
(b) Find the time it takes for $P$ to travel from $A$ to $B$.
\hfill \mbox{\textit{Edexcel Paper 3 2018 Q8 [8]}}