Edexcel M1 2022 June — Question 6 6 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeConstant acceleration vector (i and j)
DifficultyModerate -0.3 This is a straightforward SUVAT vector question requiring students to find constant acceleration from two velocity vectors, then apply it to find velocity at an intermediate time. The calculation is routine (finding acceleration components, applying v = u + at, then finding magnitude) with no conceptual challenges beyond standard M1 vector mechanics.
Spec3.02e Two-dimensional constant acceleration: with vectors

6. A particle \(P\) is moving with constant acceleration. At time \(t = 1\) second, \(P\) has velocity \(( - \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) At time \(t = 4\) seconds, \(P\) has velocity \(( 5 \mathbf { i } - 8 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\) Find the speed of \(P\) at time \(t = 3.5\) seconds.

Question 6:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((5\mathbf{i} - 8\mathbf{j}) = (-\mathbf{i} + 4\mathbf{j}) + 3\mathbf{a}\)M1 A1 Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) with \(t=3\) oe; to give equation in a only; allow u and v reversed
\(\mathbf{v} = (-\mathbf{i} + 4\mathbf{j}) + 2.5(2\mathbf{i} - 4\mathbf{j})\)M1 Equation in v only using their a and \(t = 2.5\) oe; M0 if \(\mathbf{u} = \mathbf{0}\) assumed
\(\mathbf{v} = (4\mathbf{i} - 6\mathbf{j})\)A1
Speed \(= \sqrt{4^2 + (-6)^2} = \sqrt{52} = 7.2 \ (\text{ms}^{-1})\) or betterM1 A1 Use of Pythagoras including square root; cao
OR (Ratio method):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((4-3.5)[\mathbf{v} - (-\mathbf{i}+4\mathbf{j})] = (3.5-1)[(5\mathbf{i}-8\mathbf{j}) - \mathbf{v}]\) oeM2 A1 M2 for equation in v only with correct structure; A1 correct unsimplified ratio equation
\(\mathbf{v} = (4\mathbf{i} - 6\mathbf{j})\)A1
Speed \(= \sqrt{4^2 + (-6)^2} = \sqrt{52} = 7.2 \ (\text{ms}^{-1})\) or betterM1 A1 cao
## Question 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(5\mathbf{i} - 8\mathbf{j}) = (-\mathbf{i} + 4\mathbf{j}) + 3\mathbf{a}$ | M1 A1 | Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ with $t=3$ oe; to give equation in **a** only; allow **u** and **v** reversed |
| $\mathbf{v} = (-\mathbf{i} + 4\mathbf{j}) + 2.5(2\mathbf{i} - 4\mathbf{j})$ | M1 | Equation in **v** only using their **a** and $t = 2.5$ oe; M0 if $\mathbf{u} = \mathbf{0}$ assumed |
| $\mathbf{v} = (4\mathbf{i} - 6\mathbf{j})$ | A1 | |
| Speed $= \sqrt{4^2 + (-6)^2} = \sqrt{52} = 7.2 \ (\text{ms}^{-1})$ or better | M1 A1 | Use of Pythagoras including square root; cao |

**OR (Ratio method):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(4-3.5)[\mathbf{v} - (-\mathbf{i}+4\mathbf{j})] = (3.5-1)[(5\mathbf{i}-8\mathbf{j}) - \mathbf{v}]$ oe | M2 A1 | M2 for equation in **v** only with correct structure; A1 correct unsimplified ratio equation |
| $\mathbf{v} = (4\mathbf{i} - 6\mathbf{j})$ | A1 | |
| Speed $= \sqrt{4^2 + (-6)^2} = \sqrt{52} = 7.2 \ (\text{ms}^{-1})$ or better | M1 A1 | cao |

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6. A particle $P$ is moving with constant acceleration.

At time $t = 1$ second, $P$ has velocity $( - \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$\\
At time $t = 4$ seconds, $P$ has velocity $( 5 \mathbf { i } - 8 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$\\
Find the speed of $P$ at time $t = 3.5$ seconds.

\hfill \mbox{\textit{Edexcel M1 2022 Q6 [6]}}