CAIE FP2 2011 November — Question 1 4 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeRadial and transverse acceleration
DifficultyStandard +0.3 This is a straightforward application of circular motion formulas requiring calculation of tangential acceleration (differentiation of velocity) and radial acceleration (v²/r), then combining using Pythagoras. The algebra is simple with given values, making it slightly easier than average despite being Further Maths content.
Spec6.05b Circular motion: v=r*omega and a=v^2/r6.05e Radial/tangential acceleration

1 A particle is moving in a circle of radius 2 m . At time \(t \mathrm {~s}\) its velocity is \(\left( t ^ { 2 } - 12 \right) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find the magnitude of the resultant acceleration of the particle when \(t = 4\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Tangential acceleration: \(2t = 8\)B1
Radial acceleration: \((4^2 - 12)^2/2 = 8\)B1
Magnitude: \(\sqrt{(8^2 + 8^2)} = 8\sqrt{2}\) or \(11.3\) \([\text{ms}^{-2}]\)M1 A1
Total: 4 marks
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Tangential acceleration: $2t = 8$ | B1 | |
| Radial acceleration: $(4^2 - 12)^2/2 = 8$ | B1 | |
| Magnitude: $\sqrt{(8^2 + 8^2)} = 8\sqrt{2}$ or $11.3$ $[\text{ms}^{-2}]$ | M1 A1 | |

**Total: 4 marks**

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1 A particle is moving in a circle of radius 2 m . At time $t \mathrm {~s}$ its velocity is $\left( t ^ { 2 } - 12 \right) \mathrm { m } \mathrm { s } ^ { - 1 }$. Find the magnitude of the resultant acceleration of the particle when $t = 4$.

\hfill \mbox{\textit{CAIE FP2 2011 Q1 [4]}}