Edexcel Paper 3 Specimen — Question 6 6 marks

Exam BoardEdexcel
ModulePaper 3 (Paper 3)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeVelocity from acceleration and initial conditions
DifficultyModerate -0.3 This is a straightforward integration problem requiring students to integrate acceleration to find velocity, apply initial conditions, then calculate speed at a specific time. The integration involves basic power rules (t and t^(1/2)), and finding speed from velocity components is standard. Slightly easier than average due to routine calculus and no geometric insight required.
Spec1.10c Magnitude and direction: of vectors3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration

6. At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) moves so that its acceleration \(\mathbf { a } \mathrm { m } \mathrm { s } ^ { - 2 }\) is given by $$\mathbf { a } = 5 t \mathbf { i } - 15 t ^ { \frac { 1 } { 2 } } \mathbf { j }$$ When \(t = 0\), the velocity of \(P\) is \(20 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) Find the speed of \(P\) when \(t = 4\)

Question 6:
AnswerMarks Guidance
Integrate \(\mathbf{a}\) w.r.t. timeM1 For integrating \(\mathbf{a}\) w.r.t. time (powers of \(t\) increasing by 1)
\(\mathbf{v} = \dfrac{5t^2}{2}\mathbf{i} - 10t^{\frac{3}{2}}\mathbf{j} + \mathbf{C}\) (allow omission of \(\mathbf{C}\))A1 For a correct \(\mathbf{v}\) expression without \(\mathbf{C}\)
\(\mathbf{v} = \dfrac{5t^2}{2}\mathbf{i} - 10t^{\frac{3}{2}}\mathbf{j} + 20\mathbf{i}\)A1 For a correct \(\mathbf{v}\) expression including \(\mathbf{C}\)
When \(t = 4\), \(\mathbf{v} = 60\mathbf{i} - 80\mathbf{j}\)M1 For putting \(t = 4\) into their \(\mathbf{v}\) expression
Attempt to find magnitude: \(\sqrt{(60^2 + 80^2)}\)M1 For finding magnitude of their \(\mathbf{v}\)
Speed \(= 100\) m s\(^{-1}\)A1ft ft for 100 m s\(^{-1}\), follow through on an incorrect \(\mathbf{v}\)
## Question 6:

| Integrate $\mathbf{a}$ w.r.t. time | M1 | For integrating $\mathbf{a}$ w.r.t. time (powers of $t$ increasing by 1) |
| $\mathbf{v} = \dfrac{5t^2}{2}\mathbf{i} - 10t^{\frac{3}{2}}\mathbf{j} + \mathbf{C}$ (allow omission of $\mathbf{C}$) | A1 | For a correct $\mathbf{v}$ expression without $\mathbf{C}$ |
| $\mathbf{v} = \dfrac{5t^2}{2}\mathbf{i} - 10t^{\frac{3}{2}}\mathbf{j} + 20\mathbf{i}$ | A1 | For a correct $\mathbf{v}$ expression including $\mathbf{C}$ |
| When $t = 4$, $\mathbf{v} = 60\mathbf{i} - 80\mathbf{j}$ | M1 | For putting $t = 4$ into their $\mathbf{v}$ expression |
| Attempt to find magnitude: $\sqrt{(60^2 + 80^2)}$ | M1 | For finding magnitude of their $\mathbf{v}$ |
| Speed $= 100$ m s$^{-1}$ | A1ft | ft for 100 m s$^{-1}$, follow through on an incorrect $\mathbf{v}$ |

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6. At time $t$ seconds, where $t \geqslant 0$, a particle $P$ moves so that its acceleration $\mathbf { a } \mathrm { m } \mathrm { s } ^ { - 2 }$ is given by

$$\mathbf { a } = 5 t \mathbf { i } - 15 t ^ { \frac { 1 } { 2 } } \mathbf { j }$$

When $t = 0$, the velocity of $P$ is $20 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
Find the speed of $P$ when $t = 4$

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