Edexcel M1 2018 October — Question 5 9 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2018
SessionOctober
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypePosition vector at time t (constant velocity)
DifficultyModerate -0.8 This is a straightforward M1 kinematics question requiring only direct application of the constant velocity formula r = r₀ + vt. Students substitute given values, perform basic vector arithmetic, and find a magnitude using Pythagoras. It's simpler than average A-level questions as it involves no problem-solving or conceptual challenges—just routine mechanical application of a standard formula.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors
5.
[In this question the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal vectors due east and due north respectively and position vectors are given relative to a fixed origin.]
Question 5:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Graph shape for vanB1
Graph shape for car (must cross first and end at same \(t\) value)B1
Values \(V\), 24, 12, \(T\), \(T+30\) shown with delineatorB1 (3) B0 if vertical solid lines
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{2}(T+30+30) \times 24 = 816\) OR \(\frac{1}{2} \times T \times 24 + 30 \times 24 = 816\)M1A1
\(T = 8\) sA1 (3) ALT: dist while accelerating \(= 816 - 720 = 96\); \(\left(\frac{0+24}{2}\right)T = 96\) → M1A1; \(T=8\) → A1
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{2}((T+30)+(T+18))V = 816\) OR \(\frac{1}{2} \times 12V + V(18+T) = 816\)M1A1 ft
\(V = 25.5\)A1 (3)[9]
Question 5 (part c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Dist travelled by van \(= \frac{1}{2} \times 12V + (18+T) \times V = 816\)M1A1ft Equation in \(V\) only, using 816 distance travelled by VAN, with correct structure (trapezium or triangle + rectangle). M0 if total time assumed to be 30 or 42
\(V = 25.5\)A1 cao 
# Question 5:

## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Graph shape for van | B1 | |
| Graph shape for car (must cross first and end at same $t$ value) | B1 | |
| Values $V$, 24, 12, $T$, $T+30$ shown with delineator | B1 (3) | B0 if vertical solid lines |

## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}(T+30+30) \times 24 = 816$ **OR** $\frac{1}{2} \times T \times 24 + 30 \times 24 = 816$ | M1A1 | |
| $T = 8$ s | A1 (3) | **ALT:** dist while accelerating $= 816 - 720 = 96$; $\left(\frac{0+24}{2}\right)T = 96$ → M1A1; $T=8$ → A1 |

## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}((T+30)+(T+18))V = 816$ **OR** $\frac{1}{2} \times 12V + V(18+T) = 816$ | M1A1 ft | |
| $V = 25.5$ | A1 (3)[9] | |

# Question 5 (part c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Dist travelled by van $= \frac{1}{2} \times 12V + (18+T) \times V = 816$ | M1A1ft | Equation in $V$ only, using 816 distance travelled by VAN, with correct structure (trapezium or triangle + rectangle). M0 if total time assumed to be 30 or 42 |
| $V = 25.5$ | A1 | cao |

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5.\\

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[In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are horizontal vectors due east and due north respectively and position vectors are given relative to a fixed origin.]\\

\hfill \mbox{\textit{Edexcel M1 2018 Q5 [9]}}
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