Answers to Chapter 2 problems

2.4.1 48 km h⁻¹.

2.4.1 Solution

2.4.2 v = B + 2Ct; a = 2C

At t = 3 s: s = 43.9 m; v = 25.3 m s⁻¹; a = 7.6 m s⁻².

At t = 10 s : s = 407.2 m; v = 78.5 m s⁻¹;a = 7.6 m s⁻².

2.4.2 Solution

2.4.3 v = A – 3Bt²; a = − 6Bt

At t = 0.5 s: x = 3.75 m; v = 6.5 m s⁻¹; a = − 6 m s⁻².

At t = 2.0 s : x = 0 m; v = −16 m s^-1; a = − 24 m s⁻².

2.4.3 Solution

2.4.4 (a) 0.257 m³ s⁻¹ (b) 0.161 m² s⁻¹

2.4.4 Solution

2.5.1 (a) 85 m (b) − 41 m s⁻¹ (c) 5.2 s

2.5.1 Solution

2.5.2 Either strategy, (a) or (b), would work.

2.5.2 Solution

2.5.3 (a) The stones do collide (b) 1.89 s (c) 3.29 m (d) First stone is moving downwards, the second stone upwards.

2.5.3 Solution

2.5.4 (a) 6.0 s (b)18 m s^{−1&�Բ��;} (c) 72 m.

2.5.4 Solution

2.5.5 (a)128 m (b) 5.1 s (c) 56.5 m s⁻¹

2.5.5 Solution

2.5.6 Typical graphical results:

At t = 0.5 s: a = − 5.85 m s⁻² (compares with − 6 m s^-2 from Problem 2.4.3)

At t = 2.0 s : a = − 23.2 m s⁻² (compares with − 24 m s⁻² from Problem 2.4.3)

Change of displacement = − 3.6 m (compares with − 3.75 m from Problem 2.4.3)

2.5.6 Solution

2.5.7 (a) 2 m s⁻² (b) (i) 1 m s⁻¹ (ii) 17 m s⁻¹

2.5.7 Solution

2.5.8 (a) 28.0 m s⁻¹ upwards (b) 35.7 m s⁻¹ downwards (c) 3.5 m s⁻¹ upwards

2.5.8 Solution

2.5.9 v(t) = 60 m s⁻¹ – (9.8 m s⁻²)t (a) 60 m s⁻¹ (b) 6.1 s (c) 184 m; 10.0 m s⁻²

2.5.9 Solution

2.5.10 3 m 43.2 s

2.5.10 Solution

2.6.1 96.2 m³

2.6.1 Solution

2.6.2 (a) maximum at 47.2 m; minimum at −14.4 m (b) minimum at −26.8 m s⁻¹ ; no maximum (tends to infinity)

2.6.2 Solution

2.6.3 p = 10/b + 2b; 8.94 m; l = b (a square)

2.6.3 Solution

2.7.1 (a) 1.8 rev s⁻² = 11.31 rad s⁻² (b) 3.33 s (c) 6.67 s (d) 40 rev

2.7.1 Solution

2.7.2 (a) −2.3 × 10⁻⁹ rad s^-² (b) 2620 yr (c) 0.024 s.

2.7.2 Solution

2.7.3 1.525 m; 1.82 m

2.7.3 Solution

2.7.4 −15.08 rad s−2; 2.5 s

2.7.4 Solution

2.7.5 v = 0.3wcoswt; a = − 0.3w²sinwt; 0.28 m; − 0.090 m s⁻¹; − 0.171 m s⁻².

2.7.5 Solution

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Answers to Chapter 2 problems

Understanding Physics

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