汽车在行驶过程中如果遇到危急情况,司机常要紧急刹车.从发现情况到汽车停止的距离叫做停车距离,停车距

2024-12-04 15:04:27
推荐回答(1个)
回答(1):

(1)将速度和制动距离分别对应横纵坐标,在图上描出各点,连接起来,如下图;
(2)在图上找到制动距离25m对应的速度大约是70km/h,大于50km/h,因此该司机超速;
(3)疲劳驾驶、酒后驾驶、打手机都会使司机反映迟钝或者注意力不集中,增加反应时间;超速行驶会增加反应距离,与反应时间没有关系;
(4)轮胎长期能用后会变光滑,与地面摩擦力减小,则在相同情况下,会导致制动距离变长;
(5)从图B可以看出,①发出超声波信号P1到接受到反射信号n1的时间为t1=12×

1
30
s=0.4s,
此时汽车行驶的距离S1=
1
2
Vt1=
1
2
×340m/s×0.4s=68m;
②发出超声波信号P2到接受到反射信号n2的时间为t2=9×
1
30
s=0.3s,
此时汽车行驶的距离S2=
1
2
Vt2=
1
2
×340m/s×0.3s=51m;
③所以汽车接收到P1、P2两个信号之间的时间内前进的距离为△S=S1-S2=68m-51m=17m.
④汽车运行17m的时间为汽车接收到P1、P2两个信号的时刻应分别对应于图中P1n1的中点和P2n2的中点,其间有28.5小格,即汽车接收到Pp1、P2两个信号的时间间隔为n1与n2两个信号之间的间隔,即t=28.5×
1
30
s=0.95s;
∴汽车的行驶速度V=
S
t
=
17m
0.95s
=17.9m/s.
故答案为:(1)如图所示;

(2)是;
(3)D;
(4)长;
(5)17、17.9m.

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