| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Constant acceleration vector problems |
| Difficulty | Standard +0.3 This is a straightforward constant acceleration kinematics problem using vectors. Part (a) requires finding when velocity is parallel to a given direction (setting up a ratio equation), and part (b) uses standard SUVAT with vectors. Both parts are routine applications of A-level mechanics formulas with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(\mathbf{v} = \mathbf{C} + (2\mathbf{i} - 3\mathbf{j})t\) | M1 | Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\), or integration giving form \(\mathbf{C} + (2\mathbf{i}-3\mathbf{j})t\) where C is a non-zero constant vector. M0 if u and a are reversed. Condone \(\mathbf{a} = (2\mathbf{i}+3\mathbf{j})\) |
| \(\mathbf{v} = (-\mathbf{i}+4\mathbf{j}) + (2\mathbf{i}-3\mathbf{j})t\) | A1 | Any correct unsimplified expression seen or implied |
| \(\frac{4-3T}{-1+2T} = \frac{-4}{3}\) | M1 | Correct use of ratios using a velocity vector (must use \(\frac{-4}{3}\)) to give equation in \(T\) only. M0 if they equate \(4-3T=-4\) and/or \(-1+2T=3\) |
| \(T = 8\) | A1 | Correct only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(\mathbf{s} = \mathbf{C}t + (2\mathbf{i}-3\mathbf{j})\frac{1}{2}t^2\ (+\mathbf{D})\) | M1 | Use of \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(\mathbf{a} = (2\mathbf{i}-3\mathbf{j})\), or integration giving \(\mathbf{C}t + (2\mathbf{i}-3\mathbf{j})\frac{1}{2}t^2\) where C is their non-zero constant vector from (a). Condone \(\mathbf{a}=(2\mathbf{i}+3\mathbf{j})\). Any other complete method using vector suvat equations |
| \(\mathbf{s} = (-\mathbf{i}+4\mathbf{j})t + \frac{1}{2}(2\mathbf{i}-3\mathbf{j})t^2\ (+\mathbf{D})\) | A1 | Correct unsimplified expression seen or implied |
| \(AB = \sqrt{12^2 + 8^2}\) | M1 | Use of \(t=4\) in their s (which must be a displacement vector) then Pythagoras with root sign. N.B. Beware \(4(2\mathbf{i}-3\mathbf{j})\) leading to \(\sqrt{8^2+12^2}\) is M0A0M0A0 |
| \(= 4\sqrt{13}\ (=14.422051\ldots)\) m | A1cso | Any surd form or 14 or better |
# Question 2:
## Part 2(a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $\mathbf{v} = \mathbf{C} + (2\mathbf{i} - 3\mathbf{j})t$ | M1 | Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$, or integration giving form $\mathbf{C} + (2\mathbf{i}-3\mathbf{j})t$ where C is a non-zero constant vector. M0 if **u** and **a** are reversed. Condone $\mathbf{a} = (2\mathbf{i}+3\mathbf{j})$ |
| $\mathbf{v} = (-\mathbf{i}+4\mathbf{j}) + (2\mathbf{i}-3\mathbf{j})t$ | A1 | Any correct unsimplified expression seen or implied |
| $\frac{4-3T}{-1+2T} = \frac{-4}{3}$ | M1 | Correct use of ratios using a velocity vector (must use $\frac{-4}{3}$) to give equation in $T$ only. M0 if they equate $4-3T=-4$ and/or $-1+2T=3$ |
| $T = 8$ | A1 | Correct only |
## Part 2(b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $\mathbf{s} = \mathbf{C}t + (2\mathbf{i}-3\mathbf{j})\frac{1}{2}t^2\ (+\mathbf{D})$ | M1 | Use of $\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ with $\mathbf{a} = (2\mathbf{i}-3\mathbf{j})$, or integration giving $\mathbf{C}t + (2\mathbf{i}-3\mathbf{j})\frac{1}{2}t^2$ where C is their non-zero constant vector from (a). Condone $\mathbf{a}=(2\mathbf{i}+3\mathbf{j})$. Any other complete method using vector **suvat** equations |
| $\mathbf{s} = (-\mathbf{i}+4\mathbf{j})t + \frac{1}{2}(2\mathbf{i}-3\mathbf{j})t^2\ (+\mathbf{D})$ | A1 | Correct unsimplified expression seen or implied |
| $AB = \sqrt{12^2 + 8^2}$ | M1 | Use of $t=4$ in their **s** (which must be a displacement vector) then Pythagoras with root sign. N.B. Beware $4(2\mathbf{i}-3\mathbf{j})$ leading to $\sqrt{8^2+12^2}$ is M0A0M0A0 |
| $= 4\sqrt{13}\ (=14.422051\ldots)$ m | A1cso | Any surd form or 14 or better |
---\begin{enumerate}
\item A particle, $P$, moves with constant acceleration $( 2 \mathbf { i } - 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$
\end{enumerate}
At time $t = 0$, the particle is at the point $A$ and is moving with velocity ( $- \mathbf { i } + 4 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$\\
At time $t = T$ seconds, $P$ is moving in the direction of vector ( $3 \mathbf { i } - 4 \mathbf { j }$ )\\
(a) Find the value of $T$.
At time $t = 4$ seconds, $P$ is at the point $B$.\\
(b) Find the distance $A B$.
\hfill \mbox{\textit{Edexcel Paper 3 2019 Q2 [8]}}