close
標題:
S.4 Force and motion
發問:
A drunken driver accidentally drives into a straight one-way road and meets another car which comes in opposite direction. When they are 30m apart, the drivers apply the brakes. Take direction of motion of drunken driver as positive.Drunken driver: initial velocity(kmh-1):+72 , acceleration(ms-2):-8Sober... 顯示更多 A drunken driver accidentally drives into a straight one-way road and meets another car which comes in opposite direction. When they are 30m apart, the drivers apply the brakes. Take direction of motion of drunken driver as positive. Drunken driver: initial velocity(kmh-1):+72 , acceleration(ms-2):-8 Sober driver: initial velocity(kmh-1):-54 , acceleration(ms-2):+9 1. To show cars collide. 2. How far is the position of collision from the position where drunken driver applies the brakes? 3. The velocities of cars just before collide.
最佳解答:
1. 72 km/h = 20 m/s 54 km/h = 15 m/s Consider the drunken driver's car, use equation: v^2 = u^2 + 2as with v = 0 m/s, u = 20 m/s, a = -8 m/s^2, s =? hence, 0 = 20^2 + 2.(-8).s s = 25 m Consider the sober driver's car, use the same equation of motion. 0 = (-15)^2 + 2.(9)s s = 12.5 m Hence, the total distances need to travel by the two cars before coming to rest is (25 + 12.5) m = 37.5 m, which is longer than their initial separation of 30 m. Therefore, the two cars collide. 2. Let x be the displacement travelled by the drunken driver's car before collision. Hence, displacement travelled travelled by the sober's car is -(30 -x) Use equation of motion: s = ut + (1/2)at^2 For the drunken driver's car: x = 20t + (1/2).(-8)t^2 For the sober's car: -(30-x) = -15t + (1/2).(9)t^2 Substitute the 1st equation into the 2nd one and simplify, we have 17t^2 - 70t + 60 = 0 solve for t gives t = 1.22 s or 2.9 s (rejected because of physically impossible situation) Thus distance travelled by the drunken driver's car before collision = [20 x 1.22 + (1/2) x (-8) x (1.22)^2] m = 18.45 m 3. Use equation of motion: v = u + at For the drunken driver's car: v = [20 + (-8).(1.22)] m/s = 10.24 m/s For the sober' car: v = [-15 +9.(1.22)] m/s = -4.02 m/s
其他解答:
S.4 Force and motion
發問:
A drunken driver accidentally drives into a straight one-way road and meets another car which comes in opposite direction. When they are 30m apart, the drivers apply the brakes. Take direction of motion of drunken driver as positive.Drunken driver: initial velocity(kmh-1):+72 , acceleration(ms-2):-8Sober... 顯示更多 A drunken driver accidentally drives into a straight one-way road and meets another car which comes in opposite direction. When they are 30m apart, the drivers apply the brakes. Take direction of motion of drunken driver as positive. Drunken driver: initial velocity(kmh-1):+72 , acceleration(ms-2):-8 Sober driver: initial velocity(kmh-1):-54 , acceleration(ms-2):+9 1. To show cars collide. 2. How far is the position of collision from the position where drunken driver applies the brakes? 3. The velocities of cars just before collide.
最佳解答:
1. 72 km/h = 20 m/s 54 km/h = 15 m/s Consider the drunken driver's car, use equation: v^2 = u^2 + 2as with v = 0 m/s, u = 20 m/s, a = -8 m/s^2, s =? hence, 0 = 20^2 + 2.(-8).s s = 25 m Consider the sober driver's car, use the same equation of motion. 0 = (-15)^2 + 2.(9)s s = 12.5 m Hence, the total distances need to travel by the two cars before coming to rest is (25 + 12.5) m = 37.5 m, which is longer than their initial separation of 30 m. Therefore, the two cars collide. 2. Let x be the displacement travelled by the drunken driver's car before collision. Hence, displacement travelled travelled by the sober's car is -(30 -x) Use equation of motion: s = ut + (1/2)at^2 For the drunken driver's car: x = 20t + (1/2).(-8)t^2 For the sober's car: -(30-x) = -15t + (1/2).(9)t^2 Substitute the 1st equation into the 2nd one and simplify, we have 17t^2 - 70t + 60 = 0 solve for t gives t = 1.22 s or 2.9 s (rejected because of physically impossible situation) Thus distance travelled by the drunken driver's car before collision = [20 x 1.22 + (1/2) x (-8) x (1.22)^2] m = 18.45 m 3. Use equation of motion: v = u + at For the drunken driver's car: v = [20 + (-8).(1.22)] m/s = 10.24 m/s For the sober' car: v = [-15 +9.(1.22)] m/s = -4.02 m/s
其他解答:
此文章來自奇摩知識+如有不便請留言告知
文章標籤
全站熱搜
留言列表
