Passenger car and truck

At 6 a.m., a truck left the Factory at 40 km/h. At 6:30, a car drove behind it at 70 km/h. How long, at what time, and how far from the Factory will the passenger car catch up with the truck?

Result

t =  0.6667 h
t₃ = 1:10 hh:mm
s =  46.6667 km

Step-by-step explanation:

$$\begin{gathered} t_1=6.00 \\ {t}_2=6:30=6hr30min=6+\frac{30}{60}=6.5hr=6.5 \\ {v}_1=40\text{km/h} \\ {v}_2=70\text{km/h} \\ \Delta =t_2-t_1=6.5-6=\frac{1}{2}=0.5\text{h} \\ {s}_1=s_2 \\ {v}_1\cdot (t+\Delta )=v_2\cdot t \\ 40\cdot (t+0.5)=70\cdot t \\ 30t=20 \\ t=\frac{20}{30}=0.66666667 \\ t=\frac{2}{3}\doteq 0.666667=0.6667\text{h} \end{gathered}$$

$t_3=t+\Delta =0.6667+0.5=\frac{7}{6}=1.16667=1:10\text{hh}:\text{mm}$

$s=v_2\cdot t=70\cdot 0.6667=46.6667\text{km}$

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