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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.24
Projectiles and motion with variable velocity Problem set B Answers
Skills
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- (a) 2.45 s
- (b) \(\mathbf{r} = \begin{pmatrix} 11t \\ 24t - 4.9t^2 \end{pmatrix}\)
- (c) 29.4 m
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- (a) 3.27 s
- (b) 22.9 m
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- (a) 39.5°
- (b) 20.7 m s⁻¹
- (c) 3.84 m s⁻²
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- (a) 10.0 m s⁻¹
- (b) \(\mathbf{a} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\), constant; magnitude 5.39 m s⁻²
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- (a) 9.8 m s⁻² vertically downward
- (b) 11.7 m s⁻¹
- (c) \(s = \sqrt{6^2 + (10 - 9.8t)^2}\); minimum speed 6 m s⁻¹ at t = 1.02 s
In context
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- (a) \(\mathbf{r} = \begin{pmatrix} 5t \\ 13t - 4.9t^2 \end{pmatrix}\)
- (b) It lands at t = 2.65 s, 13.3 m from the origin
- (c) Both speeds are 13.9 m s⁻¹
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- (a) \(\mathbf{v} = \begin{pmatrix} -4t \\ 6t \end{pmatrix},\ \mathbf{r} = \begin{pmatrix} -2t^2 \\ 3t^2 \end{pmatrix}\)
- (b) t = 4.16 s
- (c) Positions (0, 0), (−2, 3), (−8, 12), (−18, 27); a straight line through the origin, \(y = -1.5x\)
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- (a) \(\mathbf{v} = \begin{pmatrix} -1 + 2t \\ 6 - 3t \end{pmatrix},\ \mathbf{r} = \begin{pmatrix} -t + t^2 \\ 6t - 1.5t^2 \end{pmatrix}\)
- (b) 130°
- (c) Positions (0, 0), (0, 4.5), (2, 6), (6, 4.5); not a straight line
Stretch
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7.8 m s⁻²
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x = 0 only when t = 8, y = 0 only when t = 10 (for t > 0), never together
jmaths.xyz · AI HL · AHL 4.24 Projectiles and constant acceleration