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1. 1. ANSWERS 1. d=? vf2 = vi2 + 2ad => d = vf2 - vi2 vi = 8 2a vf = 10 => d = 102 - 82 a=6 2x6 t = =3m 2 (a) d = 400 vf2 = vi2 + 2ad => a = vf2 - vi2 vi = 25 2d vf = 15 => a = 152 - 252 a=? 2 x 400 t = = -0.5 ms-2 (b) d = (vi + vf).t d = 2d d = 2 x 400 => => 2 (vi + vf) (25 + 15) = 20 s Monday, 15 March 2010
2. 2. 3 (a) d= t = vf - vi t = 18 - 12 vf = vi + at => => vi = 12 a 1.5 vf = 18 =4s a = 1.5 t =? (b) d=? d = vf2 - vi2 => d = 182 - 122 vi = 12 2a 2 x 1.5 vf = 18 a = 1.5 d = 60 m t =4 (c) d = 100 vf2 = vi2 + 2ad => vf2 = 122 + 2 x 1.5 x 100 vi = 12 = 444 vf = ? vf =√ 444 a = 1.5 vf = 21.1 ms-1 t = Monday, 15 March 2010
3. 3. 4 (a) During this reaction time the vehicle travels with constant speed (a = 0) so: v=d => d = v.t = 12 x 0.75 = 9 m t (b) d=? d = vf2 - vi2 vf2 2 = vi + 2ad => vi = 12 2a vf = 0 d = 02 - 122 a = -2 2 x -2 t = d = 36 m (c) The dog survives 5. Consider the downward direction as positive and that the package will have the same vertical acceleration as if it were dropped from a cliff at the same altitude. (a) d = 1200 vf2 = vi2 + 2ad => vf2 = 02 + 2 X 10 X 1200 vi = 0 = 24000 vf = ? a = 10 => vf =√ 24000 t = = 155 ms-1 Monday, 15 March 2010
4. 4. 5 (b) t=? d = vit + 1at2 but vit = 0 2 t = 2d = 2 x 1200 = 15.5 s a 10 6 (a) d = 120 d = vit + 1at2 but vit = 0 vi = 0 2 vf = + t = 2d = 2 x 120 a = 10 = 4.9 s a 10 t =? (b) vf = vi + at = 0 + 10 x 4.9 = 49 ms-1 (c) t = vf - vi t = 35 - 0 d= vf = vi + at => => a 10 vi = 0 = 3.5 s vf = 35 a = 10 t =? Monday, 15 March 2010
5. 5. 7 (a) The flight path is as follows: 0 ms-1 For the 2nd half of the flight: + d= vf = vi + at - vi = 0 vf = -96 => t = vf - vi a = -10 a 96 ms-1 96 ms-1 t =? => t = 0 - 96 -10 There is the 1st half (travelling up) and the 2nd half (travelling = 9.6 s down). THE FLIGHT PATH IS SYMMETRICAL (b) We are now d= vf = vi + at analysing the vi = 96 1st half of the vf = 48 => t = vf - vi flight a = -10 a t =? => t = 48 - 96 -10 = 4.8 s Monday, 15 March 2010
6. 6. (c) We are still d= vf = vi + at analysing the vi = 96 1st half of the vf = 0 => t = vf - vi flight a = -10 a t =? => t = 0 - 96 -10 = 9.6 s (d) We are still d=? d = vf2 - vi2 vf2 2 = vi + 2ad => analysing the vi = 96 2a 1st half of the vf = 0 d = 02 - 962 flight a = -10 2 x -10 t = d = 460.8 m (e) a = -10 ms-2 Monday, 15 March 2010
7. 7. ANSWERS 1 (a) => vmw = vmg - vwg ~ ~ ~ 1.5 2.5 vmw vmw = 2.5 + 1.5 = 4 ms-1 2.5 ms-1 1.5 ms-1 Upstream (b) 4 2.5 6.5 vmg = 6.5 ms-1 downstream (c) 2.5 upstream downstream 2.5 4 4 vmg Monday, 15 March 2010
8. 8. ANSWERS 1 (a) m = man vmw = vmg - vwg => ~ ~ ~ w = water g = ground 1.5 2.5 vmw vmw = 2.5 + 1.5 = 4 ms-1 2.5 ms-1 1.5 ms-1 Upstream (b) 4 2.5 6.5 vmg = 6.5 ms-1 downstream (c) 2.5 upstream downstream 2.5 4 4 vmg Monday, 15 March 2010
9. 9. ANSWERS 1 (a) m = man vmw = vmg - vwg vmg = vmw + vwg => w = water ~ ~ ~ ~ ~ ~ g = ground 1.5 2.5 vmw vmw = 2.5 + 1.5 = 4 ms-1 2.5 ms-1 1.5 ms-1 Upstream (b) 4 2.5 6.5 vmg = 6.5 ms-1 downstream (c) 2.5 upstream downstream 2.5 4 4 vmg Monday, 15 March 2010
10. 10. ANSWERS 1 (a) m = man vmw = vmg - vwg vmg = vmw + vwg => w = water ~ ~ ~ ~ ~ ~ g = ground 1.5 2.5 vmw vmw = 2.5 + 1.5 = 4 ms-1 2.5 ms-1 1.5 ms-1 Upstream (b) vmg = vmw + vwg ~ ~ ~ 4 2.5 6.5 vmg = 6.5 ms-1 downstream (c) 2.5 upstream downstream 2.5 4 4 vmg Monday, 15 March 2010
11. 11. ANSWERS 1 (a) m = man vmw = vmg - vwg vmg = vmw + vwg => w = water ~ ~ ~ ~ ~ ~ g = ground 1.5 2.5 vmw vmw = 2.5 + 1.5 = 4 ms-1 2.5 ms-1 1.5 ms-1 Upstream (b) vmg = vmw + vwg ~ ~ ~ 4 2.5 6.5 vmg = 6.5 ms-1 downstream (c) vmg = vmw + vwg ~ ~ ~ 2.5 upstream downstream 2.5 4 4 vmg Monday, 15 March 2010
12. 12. 2 (a) vag = 20 ms-1 p = plane vpa = 60 ms-1 a = air g = ground vpg vag vpa (b) We are told that the velocity of the plane relative to the ground needs to be in the Westerly direction and asked what the pilot should do. In other words we are being asked how the plane needs to moving on the moving frame of reference (the air). This is the velocity of the plane relative to the air which is given by rearranging the equation (written in 2(a)). p v a = p v g a - v g ~ ~ ~ vpg = 60 θ (c) => tanθ = 20 -vag = 20 60 vpa => θ = 18.4o (S of E) (d) vpa = 202 + 602 = 63.2 ms-1 Monday, 15 March 2010
13. 13. 2 (a) N vag = 20 ms-1 p = plane vpa = 60 ms-1 a = air g = ground vpg vag vpa (b) We are told that the velocity of the plane relative to the ground needs to be in the Westerly direction and asked what the pilot should do. In other words we are being asked how the plane needs to moving on the moving frame of reference (the air). This is the velocity of the plane relative to the air which is given by rearranging the equation (written in 2(a)). p v a = p v g a - v g ~ ~ ~ vpg = 60 θ (c) => tanθ = 20 -vag = 20 60 vpa => θ = 18.4o (S of E) (d) vpa = 202 + 602 = 63.2 ms-1 Monday, 15 March 2010
14. 14. 2 (a) N vag = 20 ms-1 p = plane vpa = 60 ms-1 a = air g = ground vpg vag v gp = p v a + vag ~ ~ ~ vpa (b) We are told that the velocity of the plane relative to the ground needs to be in the Westerly direction and asked what the pilot should do. In other words we are being asked how the plane needs to moving on the moving frame of reference (the air). This is the velocity of the plane relative to the air which is given by rearranging the equation (written in 2(a)). p v a = p v g a - v g ~ ~ ~ vpg = 60 θ (c) => tanθ = 20 -vag = 20 60 vpa => θ = 18.4o (S of E) (d) vpa = 202 + 602 = 63.2 ms-1 Monday, 15 March 2010
15. 15. 3 (a) vwg = 1.3 b = boat w = water vbw = 2 g = ground vbg (b) vbg = vbw + vwg ~ ~ ~ 1.3 2 (c) vbg = 22 + 1.32 = 2.4 ms-1 (d) vwg = 1.3 tanθ = 1.3 => θ = tan-1(1.3/2) = 33o 2 vbw = 2 vbg => θ’ = 90 - 33 θ = 57o θ’ This is the angle relative to the bank Monday, 15 March 2010
16. 16. 3 (a) vwg = 1.3 N b = boat w = water vbw = 2 g = ground vbg (b) vbg = vbw + vwg ~ ~ ~ 1.3 2 (c) vbg = 22 + 1.32 = 2.4 ms-1 (d) vwg = 1.3 tanθ = 1.3 => θ = tan-1(1.3/2) = 33o 2 vbw = 2 vbg => θ’ = 90 - 33 θ = 57o θ’ This is the angle relative to the bank Monday, 15 March 2010