Collision

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:

$$\begin{gathered} v_1=4\text{m/s} \\ {a}_1=3{\text{m/s}}^2 \\ {v}_2=6\text{m/s} \\ {a}_2=2{\text{m/s}}^2 \\ s=240\text{m} \\ s=s_1+s_2 \\ s=(v_1+v_2)t+\frac{1}{2}{a}_1{t}^2+\frac{1}{2}{a}_2{t}^2 \\ 240=(4+6)\cdot t+0.5\cdot (3+2)\cdot t^2 \\ 240=(4+6)\cdot t+0.5\cdot (3+2)\cdot t^2 \\ -2.5t^2-10t+240=0 \\ 2.5t^2+10t-240=0 \\ a=2.5;b=10;c=-240 \\ D=b^2-4ac=10^2-4\cdot 2.5\cdot (-240)=2500 \\ D>0 \\ {t}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-10\pm \sqrt{2500}}{5} \\ {t}_{1,2}=\frac{-10\pm 50}{5} \\ {t}_{1,2}=-2\pm 10 \\ {t}_1=8 \\ {t}_2=-12 \\ t=t_1=8=8\text{s} \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} s_1=v_1\cdot t+\frac{1}{2}\cdot a_1\cdot t^2=4\cdot 8+\frac{1}{2}\cdot 3\cdot 8^2=128=128\text{m} \\ {s}_2=v_2\cdot t+\frac{1}{2}\cdot a_2\cdot t^2=6\cdot 8+\frac{1}{2}\cdot 2\cdot 8^2=112\text{m} \\ {s}_3=s_1+s_2=128+112=240\text{m} \end{gathered}$$

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