Edexcel M1 2015 January — Question 3 7 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2015
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeBearing and speed from velocity vector
DifficultyModerate -0.8 This is a straightforward mechanics question testing basic vector concepts: finding direction from components using arctangent, writing position as r₀ + vt, and solving when the position vector satisfies a simple geometric condition (north-west means equal i and j components). All parts are routine applications of standard techniques with no problem-solving insight required, making it easier than average but not trivial due to the bearing conversion and multi-step nature.
Spec1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement3.02a Kinematics language: position, displacement, velocity, acceleration

[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors directed due east and due north respectively.] A particle \(P\) is moving with constant velocity \((-6\mathbf{i} + 2\mathbf{j})\) m s\(^{-1}\). At time \(t = 0\), \(P\) passes through the point with position vector \((21\mathbf{i} + 5\mathbf{j})\) m, relative to a fixed origin \(O\).
  1. Find the direction of motion of \(P\), giving your answer as a bearing to the nearest degree. [3]
  2. Write down the position vector of \(P\) at time \(t\) seconds. [1]
  3. Find the time at which \(P\) is north-west of \(O\). [3]

Part (a)
AnswerMarks
\(\tan \alpha = 1/3 \Rightarrow \alpha \approx 18.4°\)M1 A1
Bearing is \(288°\) (nearest degree)A1 (3)
Part (b)
AnswerMarks
\(\mathbf{r} = (21\mathbf{i} + 5\mathbf{j}) + t(-6\mathbf{i} + 2\mathbf{j})\)B1 (1)
Part (c)
AnswerMarks
\(21 - 6t = -(5 + 2t)\)M1 A1
\(t = 6.5\)A1 (3)
Notes for Question 3(a):
First M1 for \(\arctan(\pm 2/\pm 6)\)
First A1 for a correct value from their expression, usually \(18.4°\) or \(71.6°\)
Second A1 for \(288\) (nearest degree)
Notes for Question 3(b):
B1 for \((21\mathbf{i} + 5\mathbf{j}) + t(-6\mathbf{i} + 2\mathbf{j})\)
Notes for Question 3(c):
M1 for equating the negative of their i-component to their j-component oe
Allow equating the components for the M mark.
First A1 for a correct equation.
Second A1 for \(t = 6.5\)
## Part (a)
$\tan \alpha = 1/3 \Rightarrow \alpha \approx 18.4°$ | M1 A1 |
Bearing is $288°$ (nearest degree) | A1 (3) |

## Part (b)
$\mathbf{r} = (21\mathbf{i} + 5\mathbf{j}) + t(-6\mathbf{i} + 2\mathbf{j})$ | B1 (1) |

## Part (c)
$21 - 6t = -(5 + 2t)$ | M1 A1 |
$t = 6.5$ | A1 (3) |

**Notes for Question 3(a):**
First M1 for $\arctan(\pm 2/\pm 6)$
First A1 for a correct value from their expression, usually $18.4°$ or $71.6°$
Second A1 for $288$ (nearest degree)

**Notes for Question 3(b):**
B1 for $(21\mathbf{i} + 5\mathbf{j}) + t(-6\mathbf{i} + 2\mathbf{j})$

**Notes for Question 3(c):**
M1 for equating the negative of their i-component to their j-component oe
Allow equating the components for the M mark.
First A1 for a correct equation.
Second A1 for $t = 6.5$

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[In this question $\mathbf{i}$ and $\mathbf{j}$ are unit vectors directed due east and due north respectively.]

A particle $P$ is moving with constant velocity $(-6\mathbf{i} + 2\mathbf{j})$ m s$^{-1}$. At time $t = 0$, $P$ passes through the point with position vector $(21\mathbf{i} + 5\mathbf{j})$ m, relative to a fixed origin $O$.

\begin{enumerate}[label=(\alph*)]
\item Find the direction of motion of $P$, giving your answer as a bearing to the nearest degree. [3]
\item Write down the position vector of $P$ at time $t$ seconds. [1]
\item Find the time at which $P$ is north-west of $O$. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2015 Q3 [7]}}