Two cars

Two cars, spaced d1 = 500 m apart, start moving against each other at time t1 = 0, the first car with acceleration a1 = 4 m/s ^ 2 and the second with acceleration a2 = 6 m/s ^ 2. At what time t2 do they meet? What distances s1 and s2 will they move from the start to the moment of meeting both cars?

Correct result:

t₂ =  10 s
s₁ =  200 m
s₂ =  300 m

Solution:

$$\begin{gathered} d_1=500\text{ m } \\ {a}_1=4{\text{m/s}}^2 \\ {a}_2=6{\text{m/s}}^2 \\ {s}_1={\displaystyle \frac{1}{2}}\cdot a_1\cdot t_2^2 \\ {s}_2={\displaystyle \frac{1}{2}}\cdot a_2\cdot t_2^2 \\ {d}_1=s_1+s_2 \\ {d}_1={\displaystyle \frac{1}{2}}\cdot a_1\cdot t_2^2+{\displaystyle \frac{1}{2}}\cdot a_2\cdot t_2^2 \\ {t}_2=\sqrt{{\displaystyle \frac{2\cdot d_1}{a_1+a_2}}}=\sqrt{{\displaystyle \frac{2\cdot 500}{4+6}}}=10\text{ s} \end{gathered}$$

$s_1={\displaystyle \frac{1}{2}}\cdot a_1\cdot t_2^2={\displaystyle \frac{1}{2}}\cdot 4\cdot 10^2=200\text{ m}$

$s_2={\displaystyle \frac{1}{2}}\cdot a_2\cdot t_2^2={\displaystyle \frac{1}{2}}\cdot 6\cdot 10^2=300\text{ m}$

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