| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find velocity by integrating acceleration |
| Difficulty | Standard +0.3 This is a straightforward Further Maths mechanics question requiring integration of vector acceleration to find velocity, then solving for when the i-component is zero, plus differentiation of position to find speed. All steps are routine applications of standard techniques with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10h 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 |
| Integrate a wrt \(t\) to obtain velocity | M1 | At least 3 terms with powers increasing by 1 (M0 if clearly just multiplying by \(t\)) |
| \(\mathbf{v} = (t - 2t^2)\mathbf{i} + \left(3t - \frac{1}{3}t^3\right)\mathbf{j}\) \((+\mathbf{C})\) | A1 | Correct expression |
| \(8\mathbf{i} - \frac{28}{3}\mathbf{j}\) (m s\(^{-1}\)) | A1 | Accept \(8\mathbf{i} - 9.3\mathbf{j}\) or better; isw if speed found |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equate i component of v to zero | M1 | Must have equation in \(t\) only (must have integrated to find velocity vector) |
| \(t - 2t^2 + 36 = 0\) | A1ft | Correct equation following through on their v; must be 3-term quadratic |
| \(t = 4.5\) (ignore incorrect second solution) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate r wrt \(t\) to obtain velocity | M1 | At least 2 terms with powers decreasing by 1 (M0 if clearly just dividing by \(t\)) |
| \(\mathbf{v} = (2t-1)\mathbf{i} + 3\mathbf{j}\) | A1 | Correct expression |
| Use magnitude to give equation in \(t\) only | M1 | Must have differentiated to find velocity; M0 if they use \(\sqrt{x^2 - y^2}\) |
| \((2t-1)^2 + 3^2 = 5^2\) | A1 | Correct equation \(\sqrt{(2t-1)^2 + 3^2} = 5\) |
| Solve 3-term quadratic for \(t\) | M1 | This M mark can be implied by a correct answer with no working |
| \(t = 2.5\) | A1 | cao |
## Question 3(i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate **a** wrt $t$ to obtain velocity | M1 | At least 3 terms with powers increasing by 1 (M0 if clearly just multiplying by $t$) |
| $\mathbf{v} = (t - 2t^2)\mathbf{i} + \left(3t - \frac{1}{3}t^3\right)\mathbf{j}$ $(+\mathbf{C})$ | A1 | Correct expression |
| $8\mathbf{i} - \frac{28}{3}\mathbf{j}$ (m s$^{-1}$) | A1 | Accept $8\mathbf{i} - 9.3\mathbf{j}$ or better; isw if speed found |
## Question 3(i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equate **i** component of **v** to zero | M1 | Must have equation in $t$ only (must have integrated to find velocity vector) |
| $t - 2t^2 + 36 = 0$ | A1ft | Correct equation following through on their **v**; must be 3-term quadratic |
| $t = 4.5$ (ignore incorrect second solution) | A1 | cao |
## Question 3(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate **r** wrt $t$ to obtain velocity | M1 | At least 2 terms with powers decreasing by 1 (M0 if clearly just dividing by $t$) |
| $\mathbf{v} = (2t-1)\mathbf{i} + 3\mathbf{j}$ | A1 | Correct expression |
| Use magnitude to give equation in $t$ only | M1 | Must have differentiated to find velocity; M0 if they use $\sqrt{x^2 - y^2}$ |
| $(2t-1)^2 + 3^2 = 5^2$ | A1 | Correct equation $\sqrt{(2t-1)^2 + 3^2} = 5$ |
| Solve 3-term quadratic for $t$ | M1 | This M mark can be implied by a correct answer with no working |
| $t = 2.5$ | A1 | cao |
---\begin{enumerate}
\item (i) At time $t$ seconds, where $t \geqslant 0$, a particle $P$ moves so that its acceleration a $\mathrm { ms } ^ { - 2 }$ is given by
\end{enumerate}
$$\mathbf { a } = ( 1 - 4 t ) \mathbf { i } + \left( 3 - t ^ { 2 } \right) \mathbf { j }$$
At the instant when $t = 0$, the velocity of $P$ is $36 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(a) Find the velocity of $P$ when $t = 4$\\
(b) Find the value of $t$ at the instant when $P$ is moving in a direction perpendicular to i\\
(ii) At time $t$ seconds, where $t \geqslant 0$, a particle $Q$ moves so that its position vector $\mathbf { r }$ metres, relative to a fixed origin $O$, is given by
$$\mathbf { r } = \left( t ^ { 2 } - t \right) \mathbf { i } + 3 t \mathbf { j }$$
Find the value of $t$ at the instant when the speed of $Q$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\hfill \mbox{\textit{Edexcel Paper 3 2020 Q3 [12]}}