Edexcel M1 — Question 4 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypePosition from velocity and initial conditions
DifficultyModerate -0.3 This is a standard M1 kinematics question involving constant velocity vectors. It requires straightforward application of s = vt, magnitude calculations, and vector addition across multiple parts. While it has several steps (12 marks total), each individual calculation is routine with no conceptual challenges or novel problem-solving required—slightly easier than average due to its mechanical nature.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration

A boy starts at the corner \(O\) of a rectangular playing field and runs across the field with constant velocity vector \((\mathbf{i} + 2\mathbf{j})\) ms\(^{-1}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors in the directions of two perpendicular sides of the field. After 40 seconds, at the point \(P\) in the field, he changes speed and direction so that his new velocity vector is \((2.4\mathbf{i} - 1.8\mathbf{j})\) ms\(^{-1}\) and maintains this velocity until he reaches the point \(Q\), where \(PQ = 75\) m. Calculate
  1. the distance \(OP\), [3 marks]
  2. the time taken to travel from \(P\) to \(Q\), [2 marks]
  3. the position vector of \(Q\) relative to \(O\). [3 marks]
Another boy travels directly from \(O\) to \(Q\) with constant velocity \((a\mathbf{i} + b\mathbf{j})\) ms\(^{-1}\), leaving \(O\) and reaching \(Q\) at the same times as the first boy.
  1. Find the values of the constants \(a\) and \(b\). [4 marks]

AnswerMarks Guidance
(a) \(P\) has p.v. \(40i + 80j\), so \(OP = \sqrt{8000} = 89.4 \text{ m}\)B1 M1 A1
(b) Speed from \(P\) to \(Q\) is \(3 \text{ m s}^{-1}\), so time \(= 25 \text{ s}\)M1 A1
(c) \(OQ = 40i + 80j + 25(2 \cdot 4i - 1 \cdot 8j) = 100i + 35j\)M1 A1 A1
(d) \(65(ai + bj) = 100i + 35j\)\(a = \frac{20}{13}, b = \frac{7}{13}\) M1 M1 A1 A1
(a) $P$ has p.v. $40i + 80j$, so $OP = \sqrt{8000} = 89.4 \text{ m}$ | B1 M1 A1

(b) Speed from $P$ to $Q$ is $3 \text{ m s}^{-1}$, so time $= 25 \text{ s}$ | M1 A1

(c) $OQ = 40i + 80j + 25(2 \cdot 4i - 1 \cdot 8j) = 100i + 35j$ | M1 A1 A1

(d) $65(ai + bj) = 100i + 35j$ | $a = \frac{20}{13}, b = \frac{7}{13}$ | M1 M1 A1 A1 | **12 marks**
A boy starts at the corner $O$ of a rectangular playing field and runs across the field with constant velocity vector $(\mathbf{i} + 2\mathbf{j})$ ms$^{-1}$, where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors in the directions of two perpendicular sides of the field. After 40 seconds, at the point $P$ in the field, he changes speed and direction so that his new velocity vector is $(2.4\mathbf{i} - 1.8\mathbf{j})$ ms$^{-1}$ and maintains this velocity until he reaches the point $Q$, where $PQ = 75$ m.

Calculate 
\begin{enumerate}[label=(\alph*)]
\item the distance $OP$, [3 marks]
\item the time taken to travel from $P$ to $Q$, [2 marks]
\item the position vector of $Q$ relative to $O$. [3 marks]
\end{enumerate}

Another boy travels directly from $O$ to $Q$ with constant velocity $(a\mathbf{i} + b\mathbf{j})$ ms$^{-1}$, leaving $O$ and reaching $Q$ at the same times as the first boy.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the values of the constants $a$ and $b$. [4 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q4 [12]}}