Two bodies 2

Two bodies start moving simultaneously from the same place in the same direction. The first body moves with uniform acceleration with an initial velocity of 4 m/s and an acceleration of 0.5 m/s², while the second body moves with uniform deceleration with an initial velocity of 10 m/s and a deceleration of 1 m/s².

Determine the time at which both bodies have travelled the same distance, and find that distance.

Final Answer:

t =  8 s
s =  48 m

Step-by-step explanation:

$$\begin{gathered} v_1=4\text{m/s} \\ {a}_1=0.5{\text{m/s}}^2 \\ {v}_2=10\text{m/s} \\ {a}_2=-1{\text{m/s}}^2 \\ {s}_1=s_2 \\ {v}_1\cdot t+\frac{1}{2}{a}_1{t}^2=v_2\cdot t+\frac{1}{2}{a}_2{t}^2 \\ 4t+0.5\cdot 0.5\cdot t^2=10\cdot t+0.5\cdot (-1)\cdot t^2 \\ 4t+0.5\cdot 0.5\cdot t^2=10\cdot t+0.5\cdot (-1)\cdot t^2 \\ 0.75t^2-6t=0 \\ a=0.75;b=-6;c=0 \\ D=b^2-4ac=6^2-4\cdot 0.75\cdot 0=36 \\ D>0 \\ {t}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{6\pm \sqrt{36}}{1.5} \\ {t}_{1,2}=\frac{6\pm 6}{1.5} \\ {t}_{1,2}=4\pm 4 \\ {t}_1=8 \\ {t}_2=0 \\ t=t_1=8=8\text{s} \end{gathered}$$

Our quadratic equation calculator calculates it.

$s=v_1\cdot t+\frac{1}{2}\cdot a_1\cdot t^2=4\cdot 8+\frac{1}{2}\cdot 0.5\cdot 8^2=48\text{m}$

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