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# 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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- 如露亦如電Lv 78 years agoFavorite Answer
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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