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發問:
A man was driving a car at 70km/h and he saw that a bay 60 m ahead started to run across the road. The man applied the brake after the car had travelled 6 m. A skid mark of 48 m was left on the road.a) What was the reaction time of the man?b) What was the average deceleration of the car when it was braked to... 顯示更多 A man was driving a car at 70km/h and he saw that a bay 60 m ahead started to run across the road. The man applied the brake after the car had travelled 6 m. A skid mark of 48 m was left on the road. a) What was the reaction time of the man? b) What was the average deceleration of the car when it was braked to stop? c) A policeman said, 'If the car had been travelling at 80km/h or faster, the boy would have been knocked down.' Verify the statement by calculating the stopping distance of the car. d) Drunk driving is a criminal offence. It has been proved that a drunk reacts more slowly. Explain how the stopping distance is affected in the case of drunk driving.
最佳解答:
(a) s = ut ------------> (s = ut +0.5at^2 with no acceleration) 6 = 70/3.6 x t t = 0.31s reaction time=0.31s (b) v^2 = u^2 + 2as 0 = (70/3.6)^2 + 2a(48) a = -3.94 ms^(-2) average deceleration= 3.94 ms^(-2) (c) stopping distance = thinking distance + braking distance thinking distance = speed x reaction time If the reaction time increases, thinking distance will increases. Thus the stopping distance will also increases. (just the idea and you can try to rewrite it) 2015-06-16 21:37:10 補充： the (c) above is (d) (c) thinking distance = (80/3.6)(0.31) = 6.88888889m 0 = (80/3.6)^2 + 2(-3.94)(s) s = 62.66842138m (braking distance = 62.66842138m) stopping distance = 6.88888889 + 62.66842138 = 69.6m > 60m ∴the boy will be knocked down ∴the statement is correct 2015-06-16 21:39:35 補充： Forgot to take 3s.f. and took 2d.p. in part (a) it should be 0.309s
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Physics 溫習題目 2發問:
A man was driving a car at 70km/h and he saw that a bay 60 m ahead started to run across the road. The man applied the brake after the car had travelled 6 m. A skid mark of 48 m was left on the road.a) What was the reaction time of the man?b) What was the average deceleration of the car when it was braked to... 顯示更多 A man was driving a car at 70km/h and he saw that a bay 60 m ahead started to run across the road. The man applied the brake after the car had travelled 6 m. A skid mark of 48 m was left on the road. a) What was the reaction time of the man? b) What was the average deceleration of the car when it was braked to stop? c) A policeman said, 'If the car had been travelling at 80km/h or faster, the boy would have been knocked down.' Verify the statement by calculating the stopping distance of the car. d) Drunk driving is a criminal offence. It has been proved that a drunk reacts more slowly. Explain how the stopping distance is affected in the case of drunk driving.
最佳解答:
(a) s = ut ------------> (s = ut +0.5at^2 with no acceleration) 6 = 70/3.6 x t t = 0.31s reaction time=0.31s (b) v^2 = u^2 + 2as 0 = (70/3.6)^2 + 2a(48) a = -3.94 ms^(-2) average deceleration= 3.94 ms^(-2) (c) stopping distance = thinking distance + braking distance thinking distance = speed x reaction time If the reaction time increases, thinking distance will increases. Thus the stopping distance will also increases. (just the idea and you can try to rewrite it) 2015-06-16 21:37:10 補充： the (c) above is (d) (c) thinking distance = (80/3.6)(0.31) = 6.88888889m 0 = (80/3.6)^2 + 2(-3.94)(s) s = 62.66842138m (braking distance = 62.66842138m) stopping distance = 6.88888889 + 62.66842138 = 69.6m > 60m ∴the boy will be knocked down ∴the statement is correct 2015-06-16 21:39:35 補充： Forgot to take 3s.f. and took 2d.p. in part (a) it should be 0.309s
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