f.4 physics 一問 20 points!!!!
2017/07/04 03:51
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f.4 physics 一問 20 points!!!!
1. A girl walks along a straight road from a point A to a point B with an average speed at 1m/s . She then returens from B to A along the same road with an average speed 2m/s . Find the average speend of the firl for the whole journey.個答案係0.67m/s , 點計? 2.A drunken driver accidentally drives into a straight ,... 顯示更多 1. A girl walks along a straight road from a point A to a point B with an average speed at 1m/s . She then returens from B to A along the same road with an average speed 2m/s . Find the average speend of the firl for the whole journey. 個答案係0.67m/s , 點計? 2.A drunken driver accidentally drives into a straight , one-way street and meets another car which comes in the opposite direction. When the cars are 30m apart, the drivers simultaneously apply the brakes. We have the following information. ( The direction of motion of the drunken driver is taken as positive ) ..................... initial velocity / km/h acceleration / m/s^2 drunken driver 72 ___________________ -8 sober driver -54____________________ 9 How far is the position of the collision from the position where the drunken driver applies the brakes ? 答案係18.4m, 點計出黎? 用中/英去做都ok, 最緊要解釋到點計, thanks.
最佳解答:
1. Let d m be the distance of the road between A and B. Total distance, s = 2d m Total time taken, t = d/1 + d/2 = 1.5d s Average speed = s/t = 2d/1.5d = 0.67 m/s 2. Let d m be the distance between the collision from the position where drunken driver applies the brakes. Consider the drunken driver: initial velocity, u = 72 km/h = 72 x (1000 m/3600 s) = 20 m/s acceleration, a = -8 m/s2 time taken, t = ? displacement s = d m s = ut + (1/2)at2 d = 20t + (1/2)(-8)t2 d = 20t - 4t2 …… (1) Consider the sober driver: initial velocity, u = -54 km/h = -54 x (1000 m/3600 s) = -15 m/s acceleration, a = 8 m/s2 time taken, t = ? displacement s = - (30 - d) m s = ut + (1/2)at2 -(30 - d) = -15t + (1/2)(9)t2 -30 + d = -15t + 4.5t2 d = 30 -15t + 4.5t2 …… (2) (1) = (2): 20t - 4t2 = 30 -15t + 4.5t2 8.5t2 - 35t + 30 = 0 t = [35 ± √(352 - 4 x 8.5 x 30)] / (2 x 8.5) t = 2.90 (rejected) or t = 1.22 Put t = 1.22 into (1): d = 20 x 1.22 - 4 x 1.222 d = 18.4 The required distance = 18.4 m
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f.4 physics 一問 20 points!!!!
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發問:1. A girl walks along a straight road from a point A to a point B with an average speed at 1m/s . She then returens from B to A along the same road with an average speed 2m/s . Find the average speend of the firl for the whole journey.個答案係0.67m/s , 點計? 2.A drunken driver accidentally drives into a straight ,... 顯示更多 1. A girl walks along a straight road from a point A to a point B with an average speed at 1m/s . She then returens from B to A along the same road with an average speed 2m/s . Find the average speend of the firl for the whole journey. 個答案係0.67m/s , 點計? 2.A drunken driver accidentally drives into a straight , one-way street and meets another car which comes in the opposite direction. When the cars are 30m apart, the drivers simultaneously apply the brakes. We have the following information. ( The direction of motion of the drunken driver is taken as positive ) ..................... initial velocity / km/h acceleration / m/s^2 drunken driver 72 ___________________ -8 sober driver -54____________________ 9 How far is the position of the collision from the position where the drunken driver applies the brakes ? 答案係18.4m, 點計出黎? 用中/英去做都ok, 最緊要解釋到點計, thanks.
最佳解答:
1. Let d m be the distance of the road between A and B. Total distance, s = 2d m Total time taken, t = d/1 + d/2 = 1.5d s Average speed = s/t = 2d/1.5d = 0.67 m/s 2. Let d m be the distance between the collision from the position where drunken driver applies the brakes. Consider the drunken driver: initial velocity, u = 72 km/h = 72 x (1000 m/3600 s) = 20 m/s acceleration, a = -8 m/s2 time taken, t = ? displacement s = d m s = ut + (1/2)at2 d = 20t + (1/2)(-8)t2 d = 20t - 4t2 …… (1) Consider the sober driver: initial velocity, u = -54 km/h = -54 x (1000 m/3600 s) = -15 m/s acceleration, a = 8 m/s2 time taken, t = ? displacement s = - (30 - d) m s = ut + (1/2)at2 -(30 - d) = -15t + (1/2)(9)t2 -30 + d = -15t + 4.5t2 d = 30 -15t + 4.5t2 …… (2) (1) = (2): 20t - 4t2 = 30 -15t + 4.5t2 8.5t2 - 35t + 30 = 0 t = [35 ± √(352 - 4 x 8.5 x 30)] / (2 x 8.5) t = 2.90 (rejected) or t = 1.22 Put t = 1.22 into (1): d = 20 x 1.22 - 4 x 1.222 d = 18.4 The required distance = 18.4 m
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