jmaths
IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.26
Circular motion Problem set A Answers
Skills
-
\(\mathbf{r} = \begin{pmatrix} 3\cos 4t \\ 3\sin 4t \end{pmatrix}\)
-
\(\mathbf{r} = \begin{pmatrix} 2\sin 3t \\ 2\cos 3t \end{pmatrix}\)
-
- (a) |r| = 5 for all t, a circle of radius 5 m
- (b) |r| = 1.5 for all t, a circle of radius 1.5 m
-
- (a) Centre \((8, −2)\), radius 1.5 m
- (b) Centre \((−3, 5)\), radius 2 m
-
- (a) Anticlockwise, 2.09 s
- (b) Clockwise, 12.6 s
- (c) Clockwise, 1.57 s
In context
-
- (a) \(\mathbf{r} = \begin{pmatrix} 2\cos 5t \\ 2\sin 5t \end{pmatrix}\)
- (b) 2 m from the centre: a circle of radius 2 m
- (c) 1.26 s
-
- (a) \(\mathbf{r} = \begin{pmatrix} 5\sin 3t \\ 5\cos 3t \end{pmatrix}\)
- (b) |r| = 5 for all t, a circle of radius 5 m
- (c) (0.706, −4.95)
-
- (a) \(\mathbf{r} = \begin{pmatrix} 15\cos \dfrac{\pi t}{15} \\ 17 + 15\sin \dfrac{\pi t}{15} \end{pmatrix}\) m
- (b) 30.0 m
- (c) 3.14 m s⁻¹
Stretch
-
- (a) \(\mathbf{r} = \begin{pmatrix} 6\cos \dfrac{t}{2} \\ 6\sin \dfrac{t}{2} \end{pmatrix}\)
- (b) |r| = 6 for all t, a circle of radius 6 m
- (c) |v| = 3 m s⁻¹, constant
jmaths.xyz · AI HL · AHL 4.26 Circular motion