OCR H240/03 2018 December — Question 7 6 marks

Exam BoardOCR
ModuleH240/03 (Pure Mathematics and Mechanics)
Year2018
SessionDecember
Marks6
TopicVectors Introduction & 2D
TypeVelocity from acceleration and initial conditions
DifficultyModerate -0.8 This is a straightforward application of constant acceleration (SUVAT) equations in vector form. Part (a) requires finding initial velocity from v = u + at, then using s = ut + ½at². Part (b) uses the initial velocity from (a) to find speed as a magnitude. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to vector manipulation and multiple steps.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors

A particle \(P\) moves with constant acceleration \((3\mathbf{i} - 5\mathbf{j})\text{m s}^{-2}\). At time \(t = 0\) seconds \(P\) is at the origin. At time \(t = 4\) seconds \(P\) has velocity \((2\mathbf{i} + 4\mathbf{j})\text{m s}^{-1}\).
  1. Find the displacement vector of \(P\) at time \(t = 4\) seconds. [2]
  2. Find the speed of \(P\) at time \(t = 0\) seconds. [4]

Part (a)
AnswerMarks Guidance
\(s = 4(2i + 4j) - \frac{1}{2}(4)^2(3i - 5j)\)M1 Attempt use of \(s = vt - \frac{1}{2}at^2\)
\(s = (-16i + 56j)m\)A1
Part (b)
AnswerMarks Guidance
\(2i + 4j = u + 4(3i - 5j)\)M1* Attempt use of \(v = u + at\)
\(\mathbf{u} = -10i + 24j\)A1
\(\mathbf{u} = \sqrt{(-10)^2 + 24^2} = 26 \text{ ms}^{-1}\)
\(= 26 \text{ ms}^{-1}\)A1
## Part (a)

$s = 4(2i + 4j) - \frac{1}{2}(4)^2(3i - 5j)$ | M1 | Attempt use of $s = vt - \frac{1}{2}at^2$ | Accept equivalent full methods using suvat equations e.g. first using $v = u + at$ to find $u$ and then using $s = ut + \frac{1}{2}at^2$

$s = (-16i + 56j)m$ | A1 |

## Part (b)

$2i + 4j = u + 4(3i - 5j)$ | M1* | Attempt use of $v = u + at$

$\mathbf{u} = -10i + 24j$ | A1 |

$|\mathbf{u}| = \sqrt{(-10)^2 + 24^2} = 26 \text{ ms}^{-1}$ | M1dep* | Attempt magnitude of their $\mathbf{u}$

$= 26 \text{ ms}^{-1}$ | A1 |
A particle $P$ moves with constant acceleration $(3\mathbf{i} - 5\mathbf{j})\text{m s}^{-2}$. At time $t = 0$ seconds $P$ is at the origin. At time $t = 4$ seconds $P$ has velocity $(2\mathbf{i} + 4\mathbf{j})\text{m s}^{-1}$.

\begin{enumerate}[label=(\alph*)]
\item Find the displacement vector of $P$ at time $t = 4$ seconds. [2]
\item Find the speed of $P$ at time $t = 0$ seconds. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/03 2018 Q7 [6]}}