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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.24
Projectiles and motion with variable velocity Problem set A Answers
Skills
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- (a) 1.84 s
- (b) \(\mathbf{r} = \begin{pmatrix} 14t \\ 18t - 4.9t^2 \end{pmatrix}\)
- (c) 16.5 m
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- (a) 2.45 s
- (b) 22.0 m
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- (a) 43.4°
- (b) 15.1 m s⁻¹
- (c) 3.12 m s⁻²
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- (a) 14.0 m s⁻¹
- (b) \(\mathbf{a} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}\), constant; magnitude 5 m s⁻²
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- (a) 9.8 m s⁻² vertically downward
- (b) 17.0 m s⁻¹
- (c) \(s = \sqrt{8^2 + (15 - 9.8t)^2}\); minimum speed 8 m s⁻¹ at t = 1.53 s
In context
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- (a) \(\mathbf{r} = \begin{pmatrix} 6t \\ 11t - 4.9t^2 \end{pmatrix}\)
- (b) It lands at t = 2.24 s, 13.5 m from the origin
- (c) Both speeds are 12.5 m s⁻¹
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- (a) \(\mathbf{v} = \begin{pmatrix} 5t \\ -2t \end{pmatrix},\ \mathbf{r} = \begin{pmatrix} 2.5t^2 \\ -t^2 \end{pmatrix}\)
- (b) t = 3.34 s
- (c) Positions (0, 0), (2.5, −1), (10, −4), (22.5, −9); a straight line through the origin, \(y = -0.4x\)
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- (a) \(\mathbf{v} = \begin{pmatrix} 6 - 3t \\ 1 + 2t \end{pmatrix},\ \mathbf{r} = \begin{pmatrix} 6t - 1.5t^2 \\ t + t^2 \end{pmatrix}\)
- (b) 80.5°
- (c) Positions (0, 0), (4.5, 2), (6, 6), (4.5, 12); not a straight line
Stretch
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3.7 m s⁻²
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x = 0 only when t = 5, y = 0 only when t = 6 (for t > 0), never together
jmaths.xyz · AI HL · AHL 4.24 Projectiles and constant acceleration