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.

Correct result:

t =  8 s
s1 =  128 m

Solution:

$v_{1}=4 \ \text{m/s} \ \\ a_{1}=3 \ \text{m/s}^2 \ \\ \ \\ v_{2}=6 \ \text{m/s} \ \\ a_{2}=2 \ \text{m/s}^2 \ \\ \ \\ s=240 \ \text{m} \ \\ \ \\ s=s_{1}+s_{2} \ \\ s=(v_{1}+v_{2})t + \dfrac{ 1 }{ 2 } a_{1} \ t^2 + \dfrac{ 1 }{ 2 } a_{2} \ t^2 \ \\ \ \\ 240=(4+6) * t + 0.5 * (3+2)*t^2 \ \\ \ \\ 240=(4+6) \cdot \ t + 0.5 \cdot \ (3+2) \cdot \ t^2 \ \\ -2.5t^2 -10t +240=0 \ \\ 2.5t^2 +10t -240=0 \ \\ \ \\ a=2.5; b=10; c=-240 \ \\ D=b^2 - 4ac=10^2 - 4\cdot 2.5 \cdot (-240)=2500 \ \\ D>0 \ \\ \ \\ t_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -10 \pm \sqrt{ 2500 } }{ 5 } \ \\ t_{1,2}=\dfrac{ -10 \pm 50 }{ 5 } \ \\ t_{1,2}=-2 \pm 10 \ \\ t_{1}=8 \ \\ t_{2}=-12 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2.5 (t -8) (t +12)=0 \ \\ \ \\ t=t_{1}=8 \ \text{s}$

Checkout calculation with our calculator of quadratic equations.

$s_{1}=v_{1} \cdot \ t + \dfrac{ 1 }{ 2 } \cdot \ a_{1} \cdot \ t^2=4 \cdot \ 8 + \dfrac{ 1 }{ 2 } \cdot \ 3 \cdot \ 8^2=128 \ \text{m} \ \\ s_{2}=v_{2} \cdot \ t + \dfrac{ 1 }{ 2 } \cdot \ a_{2} \cdot \ t^2=6 \cdot \ 8 + \dfrac{ 1 }{ 2 } \cdot \ 2 \cdot \ 8^2=112 \ \text{m} \ \\ s_{3}=s_{1}+s_{2}=128+112=240 \ \text{m}$

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