Edexcel Paper 3 2018 June — Question 6 6 marks

Exam BoardEdexcel
ModulePaper 3 (Paper 3)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypePosition from velocity and initial conditions
DifficultyStandard +0.3 This is a straightforward integration problem requiring students to integrate velocity to find position, apply initial conditions, and calculate distance between two points. The integration of t^(-1/2) is standard A-level content, and the vector arithmetic is routine. Slightly above average difficulty due to the multi-step nature and exact answer requirement, but no novel insight needed.
Spec1.10a Vectors in 2D: i,j notation and column vectors3.02f Non-uniform acceleration: using differentiation and integration

6. At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) moves in the \(x - y\) plane in such a way that its velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) is given by $$\mathbf { v } = t ^ { - \frac { 1 } { 2 } } \mathbf { i } - 4 \mathbf { j }$$ When \(t = 1 , P\) is at the point \(A\) and when \(t = 4 , P\) is at the point \(B\).
Find the exact distance \(A B\).

Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
Integrate \(\mathbf{v}\) w.r.t. timeM1 At least one power increasing by 1
\(\mathbf{r} = 2t^{\frac{3}{2}}\mathbf{i} - 2t^2\mathbf{j}\ (+\mathbf{C})\)A1 Any correct (unsimplified) expression
Substitute \(t=4\) and \(t=1\) into their \(\mathbf{r}\)M1 Must have attempted to integrate \(\mathbf{v}\); substitute both values to produce 2 vectors
\(t=4\): \(\mathbf{r} = 4\mathbf{i} - 32\mathbf{j}(+\mathbf{C})\); \(t=1\): \(\mathbf{r} = 2\mathbf{i} - 2\mathbf{j}(+\mathbf{C})\) or \((4,-32)\); \((2,-2)\)A1 These can be seen or implied
\(\sqrt{2^2 + (-30)^2}\)M1 Attempt at distance of form \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\). Must have 2 non-zero terms
\(\sqrt{904} = 2\sqrt{226}\)A1 Or any equivalent surd (exact answer needed)
# Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate $\mathbf{v}$ w.r.t. time | M1 | At least one power increasing by 1 |
| $\mathbf{r} = 2t^{\frac{3}{2}}\mathbf{i} - 2t^2\mathbf{j}\ (+\mathbf{C})$ | A1 | Any correct (unsimplified) expression |
| Substitute $t=4$ and $t=1$ into their $\mathbf{r}$ | M1 | Must have attempted to integrate $\mathbf{v}$; substitute both values to produce 2 vectors |
| $t=4$: $\mathbf{r} = 4\mathbf{i} - 32\mathbf{j}(+\mathbf{C})$; $t=1$: $\mathbf{r} = 2\mathbf{i} - 2\mathbf{j}(+\mathbf{C})$ or $(4,-32)$; $(2,-2)$ | A1 | These can be seen or implied |
| $\sqrt{2^2 + (-30)^2}$ | M1 | Attempt at distance of form $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. Must have 2 non-zero terms |
| $\sqrt{904} = 2\sqrt{226}$ | A1 | Or any equivalent surd (exact answer needed) |
6. At time $t$ seconds, where $t \geqslant 0$, a particle $P$ moves in the $x - y$ plane in such a way that its velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ is given by

$$\mathbf { v } = t ^ { - \frac { 1 } { 2 } } \mathbf { i } - 4 \mathbf { j }$$

When $t = 1 , P$ is at the point $A$ and when $t = 4 , P$ is at the point $B$.\\
Find the exact distance $A B$.

\hfill \mbox{\textit{Edexcel Paper 3 2018 Q6 [6]}}