Collision

The two bodies, whose initial distance is 240 m, move evenly against each other consistently. 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.

Result

t =  8 s
s1 =  128 m

Solution:


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s1=v1 t+12 a1 t2=4 8+12 3 82=128 m s2=v2 t+12 a2 t2=6 8+12 2 82=112 m s3=s1+s2=128+112=240 ms_{1}=v_{1} \cdot \ t + \dfrac{ 1 }{ 2 } \cdot \ a_{1} \cdot \ t^2=4 \cdot \ 8 + \dfrac{ 1 }{ 2 } \cdot \ 3 \cdot \ 8^2=128 \ \text{m} \ \\ s_{2}=v_{2} \cdot \ t + \dfrac{ 1 }{ 2 } \cdot \ a_{2} \cdot \ t^2=6 \cdot \ 8 + \dfrac{ 1 }{ 2 } \cdot \ 2 \cdot \ 8^2=112 \ \text{m} \ \\ s_{3}=s_{1}+s_{2}=128+112=240 \ \text{m}



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