2.On a particular railway,a train driverapplies the brake of the train at a yellow signal,a distance of 1.0km from ared signal,where it stops.The maximum deceleration of the train is 0.2ms–2.Assuming uniform deceleration,what is themaximum safe spe

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2.Onaparticularrailway,atraindriverappliesthebrakeofthetrainatayellowsignal,adistanceof1.0kmfromared

2.On a particular railway,a train driverapplies the brake of the train at a yellow signal,a distance of 1.0km from ared signal,where it stops.The maximum deceleration of the train is 0.2ms–2.Assuming uniform deceleration,what is themaximum safe spe
2.On a particular railway,a train driverapplies the brake of the train at a yellow signal,a distance of 1.0km from ared signal,where it stops.The maximum deceleration of the train is 0.2ms–2.
Assuming uniform deceleration,what is themaximum safe speed of the train at the yellow signal?
A.20m/s
B.40m/s
C.200m/s
D.400m/s
E.100m/s

2.On a particular railway,a train driverapplies the brake of the train at a yellow signal,a distance of 1.0km from ared signal,where it stops.The maximum deceleration of the train is 0.2ms–2.Assuming uniform deceleration,what is themaximum safe spe
A
题目说看到信号后要在1km之内停下,火车的最大加速度为0.2m/s^2,问火车的安全速度是多少
v^2=2as=2*0.2*1000(m/s)^2=400(m/s)^2
v=20m/s

选A
楼主原文有个地方打错了，最大减速度的单位应该是米每二次方秒。
由公式1/2at*t=1000，算出最大减速度时用的时间t=100秒
由公式1/2vt=1000，算出最大速度v=20米每秒。

Assume V0 is the speed at the yellow signal. apparently the speed at the red signal is 0.
The time for train moving from yellow to red T=(V0-0)/0.2=0.2V0.
Then 0.5(V0+0)*0.2V0=1000;
Thus,V0=100m/s, which is the maximum safe speed.

可以

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