Anonymous
Anonymous asked in 科學其他:科學 · 8 years ago

普物運動學一題

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.

(a) Where and when do the tow cars collide?

1 Answer

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  • 8 years ago
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    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+16tB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

    相撞時Xa=Xb

    -t^2+16t = -2t^2 + 28t -35

    =>t^2-12t+35 = 0

    =>(t-5)(t-7)=0

    =>t=5 秒(已相撞不必再算t=7秒)A車位置Xa= -t^2+16t =-25+16*5=55m

    (B車行走距離Sb=24+24*(4)+0.5*(-4)*(4)^2=88m)

    Source(s): 如是我聞
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