CAIE FP2 2018 June — Question 1 3 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2018
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeRadial and transverse acceleration
DifficultyStandard +0.3 This is a straightforward application of standard circular motion formulas (radial acceleration = v²/r, transverse acceleration = dv/dt) with simple differentiation and substitution. The question requires recall of two formulas and basic calculus, but involves no problem-solving or conceptual challenges beyond textbook exercises.
Spec6.05e Radial/tangential acceleration

1 A particle \(P\) is moving in a fixed circle of radius 0.8 m . At time \(t \mathrm {~s}\) its velocity is \(\left( t ^ { 2 } - t + 2 \right) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find the magnitudes of the radial and the transverse components of the acceleration of \(P\) when \(t = 2\). Radial component
Transverse component \(\_\_\_\_\)

Question 1:
AnswerMarks Guidance
\(a_R = (2^2 - 2 + 2)^2 / 0.8 = 4^2 / 0.8 = 20 \text{ [m s}^{-2}\text{]}\)M1 A1 Find radial acceleration \(a_R\) at \(t = 2\) from \(v^2/r\)
\(a_T = 2t - 1 = 3 \text{ [m s}^{-2}\text{]}\)B1 Find transverse acceleration \(a_T\) at \(t = 2\) by differentiation
Total: 3
**Question 1:**

$a_R = (2^2 - 2 + 2)^2 / 0.8 = 4^2 / 0.8 = 20 \text{ [m s}^{-2}\text{]}$ | M1 A1 | Find radial acceleration $a_R$ at $t = 2$ from $v^2/r$

$a_T = 2t - 1 = 3 \text{ [m s}^{-2}\text{]}$ | B1 | Find transverse acceleration $a_T$ at $t = 2$ by differentiation

**Total: 3**

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1 A particle $P$ is moving in a fixed circle of radius 0.8 m . At time $t \mathrm {~s}$ its velocity is $\left( t ^ { 2 } - t + 2 \right) \mathrm { m } \mathrm { s } ^ { - 1 }$. Find the magnitudes of the radial and the transverse components of the acceleration of $P$ when $t = 2$.

Radial component\\

Transverse component $\_\_\_\_$\\

\hfill \mbox{\textit{CAIE FP2 2018 Q1 [3]}}