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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.29
Circular motion with changing speed Problem set A Answers
Skills
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\(|\mathbf{r}| = 4.5\): a circle of radius 4.5 m
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\(\mathbf{v} = \begin{pmatrix}-0.8t\sin(0.5t^2)\\0.8t\cos(0.5t^2)\end{pmatrix}\) m s⁻¹
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\(\mathbf{v}\cdot\mathbf{r} = 0\), so they are perpendicular
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\(\mathbf{a} = \begin{pmatrix}-0.44\sin(0.2t^2) − 0.176t^2\cos(0.2t^2)\\0.44\cos(0.2t^2) − 0.176t^2\sin(0.2t^2)\end{pmatrix}\) m s⁻²
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\(t = 5.39\) s
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10 complete revolutions
In context
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- (a) \(|\mathbf{r}| = 2.4\), a circle of radius 2.4 m
- (b) \(\mathbf{v} = \begin{pmatrix}-0.24t\sin(0.05t^2)\\0.24t\cos(0.05t^2)\end{pmatrix}\) m s⁻¹
- (c) \(\mathbf{v}\cdot\mathbf{r} = 0\)
- (d) \(\mathbf{a} = \begin{pmatrix}-0.24\sin(0.05t^2) − 0.024t^2\cos(0.05t^2)\\0.24\cos(0.05t^2) − 0.024t^2\sin(0.05t^2)\end{pmatrix}\) m s⁻²
- (e) \(t = 20.4\) s
- (f) 3 complete revolutions
Stretch
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- (a) \(|\mathbf{r}| = 1.6\), a circle of radius 1.6 m
- (b) \(\mathbf{v} = \begin{pmatrix}0.96t\cos(0.3t^2)\\−0.96t\sin(0.3t^2)\end{pmatrix}\) 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