一道物理题
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S1\uff0cS2\u90fd\u90fd\u65ad\u5f00\u65f6\uff0cR1R2R3\u4e32\u8054\uff08\u60c5\u51b5\u4e00\uff09\uff0c\u53ea\u95ed\u5408S2\u65f6\uff0cR1R2\u4e32\u8054\uff08\u60c5\u51b5\u4e8c\uff09\uff0c\u53ea\u95ed\u5408S1\u65f6\uff0c\u7535\u8def\u4e2d\u53ea\u5269\u4e0bR1\uff08\u60c5\u51b5\u4e09\uff09\u3002\u5bf9\u6bd4\u60c5\u51b5\u4e00\u548c\u60c5\u51b5\u4e09\uff0c\u6839\u636ep=I^2R\uff0c\u53ef\u77e5I1/I3=1/6\uff0c\u6545(R1+R2+R3)/R1=6/1(\u4e00\u5f0f)\u3002\u5355\u770b\u60c5\u51b5\u4e00\uff0c\u6839\u636ep=I^2R\uff0cp1/p3=R1/R3\uff0c\u53c8\u56e0\u4e3ap1=1w\uff0c\u6240\u4ee5p3=R3/R1\u3002\u5355\u770b\u60c5\u51b5\u4e8c\uff0c\u6839\u636ep=I^2R\uff0cI=U/R\uff0c\u53ef\u5f97(U/(R1+R2))^2R2=8(\u4e8c\u5f0f)\u3002\u5355\u770b\u60c5\u6982\u51b5\u4e09\uff0c\u6839\u636ep=U^2/R\uff0c\u53ef\u5f97U^2/R1=1(\u4e09\u5f0f)\u3002.\u7531\u4e00\u5f0f\u53ef\u5f97\uff0cR2=5R1-R3(\u56db\u5f0f)\uff0c\u5c06\u56db\u5f0f\u4ee3\u5165\u4e8c\u5f0f\u5e76\u4e0e\u4e09\u5f0f\u505a\u6bd4\uff0c\u53ef\u5f9727(R1)^2-15R1R3+2(R3)^2=0(\u4e94\u5f0f)\uff0c\u4e94\u5f0f\u4e24\u8fb9\u540c\u9664\u4ee5(R1)^2\uff0c\u53ef\u6c42\u53ef\u5f97R3/R1=3\u62164.5\u3002.\u518d\u6839\u636eR1>R2\uff0c\u4ee5\u53ca\u4e00\u5f0f\uff0c\u53ef\u820d\u53bbR3/R1=3\u3002\u6545R3/R1=4.5\uff0c\u5373P3=4.5w\u3002
解析：由题意可知：设汽车的加速度为a,初速度为V0 ,两根电线杆之间的距离S=50m,汽车通过第一个间隔用了t=5s,通过第二个间隔用了t =4s
第一个50m
S=V0t+1/2at² =50m (1)
当通过第二个间隔时，汽车的位移为S1=50m+50m=100m
行驶位移100m时所用的时间t2=t+t1=9s (2)
S1=V0t2+1/2at2²=100m (3)
由（1）（2）（3） 解得
a= 0.56m/s²
tretre
设经过前面一根电线杆时速度是V，加速度是a ，则
经第一个间隔，50＝V*5＋a*5^2 /2
经全部两个间隔，50＋50＝V*（5＋4）＋a*（5＋4）^2 /2
两式联立得 加速度是 a＝5/9＝0.56 m/s^2
5/9m/s²
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