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From cities A and B, which are 50 km apart, two cars set off in the same direction simultaneously with average speeds of 80 km/h and 120 km/h. How long will the faster car reach, the slower one, and at what distance from city A?

Final Answer:

t =  0.25 h
s₁ =  20 km

Step-by-step explanation:

$$\begin{gathered} s=50\text{km} \\ {v}_1=80\text{km/h} \\ {v}_2=120\text{km/h} \\ s=v\cdot t \\ t=\frac{s}{v_1+v_2}=\frac{50}{80+120}=\frac{1}{4}\text{h}=0.25\text{h} \end{gathered}$$

$s_1=v_1\cdot t=80\cdot \frac{1}{4}=\frac{80\cdot 1}{4}=\frac{80}{4}=20\text{km}$

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