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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.29
Circular motion with changing speed Problem set B Answers
Skills
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\(|\mathbf{r}| = 0.7\): a circle of radius 0.7 m
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\(\mathbf{v} = \begin{pmatrix}-0.7t\sin(0.25t^2)\\0.7t\cos(0.25t^2)\end{pmatrix}\) m s⁻¹
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\(\mathbf{v}\cdot\mathbf{r} = 0\), so they are perpendicular
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\(\mathbf{a} = \begin{pmatrix}-0.18\sin(0.1t^2) − 0.036t^2\cos(0.1t^2)\\0.18\cos(0.1t^2) − 0.036t^2\sin(0.1t^2)\end{pmatrix}\) m s⁻²
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\(t = 7.89\) s
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9 complete revolutions
In context
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- (a) \(|\mathbf{r}| = 0.9\), a circle of radius 0.9 m
- (b) \(\mathbf{v} = \begin{pmatrix}-0.36t\sin(0.2t^2)\\0.36t\cos(0.2t^2)\end{pmatrix}\) m s⁻¹
- (c) \(\mathbf{v}\cdot\mathbf{r} = 0\)
- (d) \(\mathbf{a} = \begin{pmatrix}-0.36\sin(0.2t^2) − 0.144t^2\cos(0.2t^2)\\0.36\cos(0.2t^2) − 0.144t^2\sin(0.2t^2)\end{pmatrix}\) m s⁻²
- (e) \(t = 9.13\) s
- (f) 2 complete revolutions
Stretch
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- (a) \(|\mathbf{r}| = 2\), a circle of radius 2 m
- (b) \(\mathbf{v} = \begin{pmatrix}1.2t\cos(0.3t^2)\\−1.2t\sin(0.3t^2)\end{pmatrix}\) m s⁻¹
- (c) \(\mathbf{v}\cdot\mathbf{r} = 0\)
- (d) \(t = 5\) s
jmaths.xyz · AI HL · AHL 4.29 Exam practice: circular motion with changing speed