一輛自行車從A點(diǎn)由靜止出發(fā)做勻加速直線運(yùn)動(dòng),用7s時(shí)間通過一座長BC=14m的平橋,過橋后的速度是3m/s,求:                                                                                                                                

(1)它剛開上橋頭B時(shí)的速度vB有多大?                                                                  

(2)橋頭B與出發(fā)點(diǎn)相距多遠(yuǎn)?                                                                                 

                                                                                                                                    


(1)在過橋過程中的平均速度==2m/s

勻變速直線運(yùn)動(dòng)的平均速度

故:vB=1m/s

(2)勻變速直線運(yùn)動(dòng)的加速度a===

根據(jù)速度位移公式v02=2ax,代入數(shù)據(jù)得,

     x=m=1.75m

故橋頭與出發(fā)點(diǎn)間的距離為1.75m.

答:(1)它剛開上橋頭B時(shí)的速度vB為1m/s;

(2)橋頭B與出發(fā)點(diǎn)相距1.75m.


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