一輛值勤的*車停在平直公路上的A點，當*員發現從他旁邊以v＝9m/s的速度勻速駛過的貨車有違章行為時，決定前去...

問題詳情：

一輛值勤的*車停在平直公路上的A點，當*員發現從他旁邊以v＝9m/s的速度勻速駛過的貨車有違章行為時，決定前去追趕。*車啟動時貨車已運動到B點，A、B兩點相距x0=48 m，*車從A點由靜止開始向右做勻加速運動，到達B點後開始做勻速運動．已知*車從開始運動到追上貨車所用的時間t＝32 s，求：

(1) *車加速運動過程所用的時間t1和加速度a的大小

(2) *車追上貨車之前的最遠距離x

【回答】

(1)設*車加速過程所用的時間為t1，加速度大小為a

則x0＝at·························································································· (1分)

at1(t－t1)＝v t···························································································· (2分)

得t1＝8 s·································································································· (1分)

a＝1.5m/s2····························································································· (1分)

(2)設經t0*車與貨車共速，此時*車追上貨車之前最遠

t0＝=6s································································································· (1分)

t0時間內貨車運動的距離x1＝v t0=54m······················································ (1分)

t0時間內*車運動的距離x2＝t0=27 m···················································· (1分)

此時相距x= x0+ x1- x2=75 m······································································ (2分)

知識點：勻變速直線運動的研究單元測試

題型：計算題