The two bodies, whose initial distance is 240 m, consistently move evenly against each other. The first body has an initial velocity of 4 m/s and an acceleration of 3 m/s2. The second body has an initial speed of 6 m/s and an acceleration of 2 m/s2. Find the time of collision between the bodies and the collision distance from the initial position of the first body.

Correct answer:

t =  8 s
s1 =  128 m

Step-by-step explanation:

v1=4 m/s a1=3 m/s2  v2=6 m/s a2=2 m/s2  s=240 m  s = s1+s2 s = (v1+v2)t + 21 a1 t2  + 21 a2 t2  240=(4+6) t+0.5 (3+2) t2  240=(4+6) t+0.5 (3+2) t2 2.5t210t+240=0 2.5t2+10t240=0  a=2.5;b=10;c=240 D=b24ac=10242.5(240)=2500 D>0  t1,2=2ab±D=510±2500 t1,2=510±50 t1,2=2±10 t1=8 t2=12  t=t1=8=8 s

Our quadratic equation calculator calculates it.

s1=v1 t+21 a1 t2=4 8+21 3 82=128=128 m s2=v2 t+21 a2 t2=6 8+21 2 82=112 m s3=s1+s2=128+112=240 m

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