OCR MEI Further Mechanics Major Specimen — Question 1 4 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (vectors)
TypeMagnitude of acceleration at given time
DifficultyModerate -0.3 This is a straightforward kinematics question requiring two differentiations and then calculating a magnitude. The calculus is routine (polynomial differentiation), and the method is standard for Further Maths mechanics. It's slightly easier than average A-level due to being purely procedural with no problem-solving or conceptual challenges.
Spec1.07a Derivative as gradient: of tangent to curve1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration

A particle P has position vector \(\mathbf{r}\) m at time \(t\) s given by \(\mathbf{r} = (t^3 - 3t^2)\mathbf{i} - (4t^2 + 1)\mathbf{j}\) for \(t \geq 0\). Find the magnitude of the acceleration of P when \(t = 2\). [4]

Question 1:
AnswerMarks
1v = (3t2 (cid:16)6t)i – 8t j
a = (6t – 6) i – 8 j
When t = 2, a = 6i – 8j
AnswerMarks
Magnitude of acceleration = 10 ms-2M1
M1
M1
A1
AnswerMarks
[4]1.1a
1.1
1.1
AnswerMarks
1.1Differentiate
Differentiate
Substitute t = 2
AnswerMarks Guidance
14 0
Question 1:
1 | v = (3t2 (cid:16)6t)i – 8t j
a = (6t – 6) i – 8 j
When t = 2, a = 6i – 8j
Magnitude of acceleration = 10 ms-2 | M1
M1
M1
A1
[4] | 1.1a
1.1
1.1
1.1 | Differentiate
Differentiate
Substitute t = 2
1 | 4 | 0 | 0 | 0 | 4
A particle P has position vector $\mathbf{r}$ m at time $t$ s given by $\mathbf{r} = (t^3 - 3t^2)\mathbf{i} - (4t^2 + 1)\mathbf{j}$ for $t \geq 0$.

Find the magnitude of the acceleration of P when $t = 2$. [4]

\hfill \mbox{\textit{OCR MEI Further Mechanics Major  Q1 [4]}}