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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.26
Circular motion Problem set B Answers
Skills
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\(\mathbf{r} = \begin{pmatrix} 2\cos 7t \\ 2\sin 7t \end{pmatrix}\)
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\(\mathbf{r} = \begin{pmatrix} 5\sin 4t \\ 5\cos 4t \end{pmatrix}\)
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- (a) |r| = 9 for all t, a circle of radius 9 m
- (b) |r| = 2.5 for all t, a circle of radius 2.5 m
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- (a) Centre \((−5, 6)\), radius 2.4 m
- (b) Centre \((7, −1)\), radius 4 m
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- (a) Anticlockwise, 3.14 s
- (b) Clockwise, 18.8 s
- (c) Clockwise, 1.26 s
In context
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- (a) \(\mathbf{r} = \begin{pmatrix} 3\cos 6t \\ 3\sin 6t \end{pmatrix}\)
- (b) 3 m from the centre: a circle of radius 3 m
- (c) 1.05 s
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- (a) \(\mathbf{r} = \begin{pmatrix} 6\sin 4t \\ 6\cos 4t \end{pmatrix}\)
- (b) |r| = 6 for all t, a circle of radius 6 m
- (c) (−4.54, −3.92)
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- (a) \(\mathbf{r} = \begin{pmatrix} 2.4\cos \dfrac{\pi t}{4} \\ 3.1 − 2.4\sin \dfrac{\pi t}{4} \end{pmatrix}\) m
- (b) 1.40 m
- (c) 1.88 m s⁻¹
Stretch
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- (a) \(\mathbf{r} = \begin{pmatrix} 8\cos \dfrac{t}{4} \\ 8\sin \dfrac{t}{4} \end{pmatrix}\)
- (b) |r| = 8 for all t, a circle of radius 8 m
- (c) |v| = 2 m s⁻¹, constant
jmaths.xyz · AI HL · AHL 4.26 Circular motion