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參芳答案
=10
專題一直線運(yùn)動
位移x2=
2a12X2m=25m
電梯加連到最大述度的時問，==0
8=5
1.【解析】汽車的速度v=21.6km/h=6m/s,
汽車在前t=0.38十0.78=1.0s內(nèi)做勻速直線運(yùn)動，位
w2=10
移為：x1=t=6X1m=6m
電梯減速運(yùn)動的位移x,一2：=2X]m=50m
隨后汽車微減速運(yùn)動，位移為：x,=2a=2X5m=3.6m
電梯減速運(yùn)動的時間1，=”=10
a,=T8=l08
所以該ETC通連的長度為：L=x1十x2=6m十3,6m=
勻速運(yùn)動的時間1，=工一x)一x-84-25-50
8=0.9s
10
9.6m
【答案】9.6m
則總時間t=t1十t2十tg=15.9s
2.【解析】(1)設(shè)泥石流到達(dá)坡底的時間為1，速度為
根據(jù)位移公式有，=十1G
(3)設(shè)最大連度為，則整個過程受4，'十-'
由速度公式有1=v十11
+受4，'=h
代入數(shù)據(jù)得v=8m/s
其中h=84m,t=24s
(2)泥石流到達(dá)坡底時，汽車速度=ag(t1一t)
又a1t1'=azt'=um
汽車位移x=2a:(4一)廣
解得vm=4m/s
【答案】(1)16m(2)15.98(3)4m/s
當(dāng)汽車的速度與泥石流速度相等時，泥石流與汽車相距
6.【解析】(1)當(dāng)兩車速度相等時，距離最遠(yuǎn)，設(shè)t1時兩車
最近
設(shè)泥石流到達(dá)坡底后汽車又加速時間為2，故有1一
速度相等，則有1==10
108=13
4:t:=v十a(chǎn):l:
泥石流水平位移=4一，
兩車相距最遠(yuǎn)的距離為△x=4十x,一之a(chǎn)t行=10X】
汽車位移xA=十之a(chǎn),號
1
m+15m-號×10×12m=20m
相距最近的距離x=x。十xA一x無
(2)設(shè)A車經(jīng)過1時間追上B車，則有，1+x,=號
由以上各式解得x=6m
代入數(shù)據(jù)可得1一2一3=0
【答案】(1)8m/s(2)6m
解得t=38或t=一1s(含去)
3.【解析】(1)設(shè)潛水器加速階段的位移為x1,運(yùn)動時間
此時A車的速度為v=at=30m/s
為，末速度為，減速階段的位移為x,運(yùn)動時間為，
則由勻變速直線運(yùn)動規(guī)律有v=21x1
(3)A車停下來需要的時間為1=”=30
41=48=7.5s
0-w2=2(-42）x2
假設(shè)A車停止前兩車再次相遇，設(shè)兩車相遇后再次相遇
又x1十x:=3000m
聯(lián)主代入數(shù)據(jù)可得x1=1000mv=20m/s
經(jīng)過時間為4，則有=，一2a1號
(2)根據(jù)勻變速直線運(yùn)動速度一時間公式可得
代入數(shù)據(jù)解得t2=10s>t或t。=0(舍去)
v=at
則兩車再次相遇時A車已經(jīng)停止，假設(shè)不成立；設(shè)兩車
0=v-d2t2
又tg=t+g
相遇后再次相逼經(jīng)過時間為4，則有，=受
聯(lián)立整理代入數(shù)據(jù)解得t=3003,
解得t,=11.25s
【答案】(1)1000m(2)300s
【答案】(1)20m(2)3s,30m/3(3)11.25s
4.【解析】(1)根據(jù)平均建度定義口=，解得汽車在演區(qū)
7.【解析】(1)設(shè)A車用時t追上B車，對A車，xA=t
1
ant
間運(yùn)動的最短總時間1=于=0h=0.19h=684。
對B車，x8=Ut
(2)設(shè)勻加速行駛的時間為t1,勻速行駛的時間為t,汽
相過時有x =xB十x
車在區(qū)間測速起點(diǎn)的速度為，勻速行駛的速度為凸，
解得=18,2=78
區(qū)間測速終點(diǎn)的最高速度為，總位移為x,勻加速行
顯然1為A車追上B車，由于12=7s>1=6s故4，
歌的位移工=嗎十墜1，勻追行駛的位移，=，4，勻減
2
為A車停下后祓B(yǎng)車追上，
速行駛的位移x,=心十”(1一1，一，)
2
十x
設(shè)從開始到A車被B車追上月時為1s,則：=2a
總位移x=x1十x2十x
得1=7.25s
聯(lián)主解得：v=72km/小
所以△1=1g一t1,解得△1=6.25s
【答案】(1)684s(2)72km/h
(2)設(shè)當(dāng)A車與B車速度相等用時為t,則A一at,=
5.【解析】(1)若上升過程中最大速度為8m/s,啟動到最
U:,t4=4s
大建度位移，氣若-2式3m=16m
1
則此過程中A車位移r'A=一之“，B車位移t'B
(2)若上升過程中最大速度為10m/s,啟動到最大遂度
=B1
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