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匿名使用者 發問時間: 科學其他:科學 · 8 年前

普物運動學一題-c

Car A is moving at a speed of 16 m/s. Car B is approaching from behind with a speed of 24 m/s. When the driver of car A sees a stop sign ahead and starts to decelerate woth an acceleration of -2 m/s^2, car B is 33m behind. Car B starts to decelerate with an acceleration of -4 m/s^2 one second later.

(c) Sketch the positions as functions of time for these two objects after car A starts to decelerate.

1 個解答

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  • 8 年前
    最佳解答

    設A車行走 t 秒,並假設以A車t=0,位置為0A車行走距離Sa=16*t+0.5*(-2)*t^2

    =>A車位置Xa=0 + [ 16t-t^2]= -t^2+16t B車行走( t-1) 秒,B車t=0,位置為-33m(速度t=0~1秒等速24m/s)

    B車行走距離Sb=24+ 24*(t-1)+0.5*(-4)*(t-1)^2=> B車位置Xb =-33 + [24t-2(t^2-2t+1)]= -2t^2+28t-35

    參考資料: 如是我聞
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