Physics 溫習題目 2

[匿名]

2015-06-17 2:42 am

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.

回答 (1)

用戶45314

2015-06-17 4:42 am

✔ 最佳答案

(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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