Edexcel M1 — Question 5 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeClosest approach of two objects
DifficultyStandard +0.3 This is a standard M1 closest approach problem with straightforward vector arithmetic. Students find displacement, minimize distance using calculus (or completing the square), requiring 2-3 routine steps with no geometric insight needed. Slightly easier than average due to its mechanical, textbook nature.
Spec1.10e Position vectors: and displacement1.10f Distance between points: using position vectors3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area

5. Two dogs, Fido and Growler, are playing in a field. Fido is moving in a straight line so that at time \(t\) his position vector relative to a fixed origin, \(O\), is given by \([ ( 2 t - 3 ) \mathbf { i } + t \mathbf { j } ]\) metres. Growler is stationary at the point with position vector \(( 2 \mathbf { i } + 5 \mathbf { j } )\) metres, where \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal perpendicular unit vectors.
  1. Find the displacement vector of Fido from Growler in terms of \(t\).
  2. Find the value of \(t\) for which the two dogs are closest.
  3. Find the minimum distance between the two dogs.

AnswerMarks Guidance
(a) disp. of F rel to G \(= [(2t - 3) - 2]\mathbf{i} + [(t - 5) - 5]\mathbf{j} = (2t - 5)\mathbf{i} + (t - 5)\mathbf{j}\)M1 A1
(b) \(d^2 = (2t - 5)^2 + (t - 5)^2\)M1
\(= 4t^2 - 20t + 25 + t^2 - 10t + 25 = 5t^2 - 30t + 50\)M1 A1
\(= 5(t^2 - 6t + 10) = 5[(t - 3)^2 + 1]\)M2
min. \(d^2\) (and hence \(d\)) when \(t = 3\)A1
(c) when \(t = 3, d^2 = 5\)M1 A1
dist. \(= \sqrt{5} = 2.24 \text{ m}\)A1 (3sf)
**(a)** disp. of F rel to G $= [(2t - 3) - 2]\mathbf{i} + [(t - 5) - 5]\mathbf{j} = (2t - 5)\mathbf{i} + (t - 5)\mathbf{j}$ | M1 A1 |

**(b)** $d^2 = (2t - 5)^2 + (t - 5)^2$ | M1 |
$= 4t^2 - 20t + 25 + t^2 - 10t + 25 = 5t^2 - 30t + 50$ | M1 A1 |
$= 5(t^2 - 6t + 10) = 5[(t - 3)^2 + 1]$ | M2 |
min. $d^2$ (and hence $d$) when $t = 3$ | A1 |

**(c)** when $t = 3, d^2 = 5$ | M1 A1 |
dist. $= \sqrt{5} = 2.24 \text{ m}$ | A1 | (3sf) | (11)
5. Two dogs, Fido and Growler, are playing in a field. Fido is moving in a straight line so that at time $t$ his position vector relative to a fixed origin, $O$, is given by $[ ( 2 t - 3 ) \mathbf { i } + t \mathbf { j } ]$ metres. Growler is stationary at the point with position vector $( 2 \mathbf { i } + 5 \mathbf { j } )$ metres, where $\mathbf { i }$ and $\mathbf { j }$ are horizontal perpendicular unit vectors.
\begin{enumerate}[label=(\alph*)]
\item Find the displacement vector of Fido from Growler in terms of $t$.
\item Find the value of $t$ for which the two dogs are closest.
\item Find the minimum distance between the two dogs.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q5 [11]}}