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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-b

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.

(b) Find the minimun acceleration of car B so that a collision can be avoided.

1 個解答

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  • 8 年前
    最佳解答

    設A車行走 t 秒A車行走距離Sa=16*t+ 0.5*(-2)*t^2

    =>A車位置Xa=0 + [ 16t-t^2 ]= -t^2+16tB車行走( t-1) 秒(速度t=0~1秒等速Vo=V1=24m/s)

    B車行走距離Sb=24+ 24*(t-1)+ 0.5*a*(t-1)^2

    =24t+0.5a(t-1)^2

    將相撞時Xb=Xa,Vb=Va

    24t+0.5a(t-1)^2= (-t^2+16t)+33

    =>a=-(2t^2+16t-66)/(t-1)^2

    Vb=Va =>24+a*(t-1)=16-2*t

    =>24-[(2t^2+16t-66)/(t-1)^2]*(t-1)=16-2*t

    =>24(t-1)-(2t^2+16t-66)= (16-2t)(t-1)

    =>t=5.8

    代入

    24+a*(t-1)=16-2*t

    =>24+4.8a=4.4

    a= -4.083 m/s2 -----min

    參考資料: 如是我聞
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