The factory 3

The factory is 18 km away from the city. A cyclist leaves the factory for the city, and half an hour after him from the same place, a motorcyclist arrives 6 minutes before the cyclist in the city. Find the speed of each, knowing that the speed of the cyclist is two times smaller than that of the motorcyclist.

Correct answer:

v₁ =  15 km/h
v₂ =  30 km/h

Step-by-step explanation:

$$\begin{gathered} s=18\text{km} \\ {t}_1=1/2=\frac{1}{2}=0.5\text{h} \\ {t}_2=6\text{min}\to \text{h}=6:60\text{h}=0.1\text{h} \\ \Delta =t_1+t_2=0.5+0.1=\frac{3}{5}=0.6\text{h} \\ s=s_1=s_2 \\ s=v_1\cdot t \\ s=v_2\cdot (t-\Delta ) \\ {v}_1=v_2/2 \\ s=v_1\cdot t \\ s=2\cdot v_1\cdot (t-\Delta ) \\ s=2\cdot v_1\cdot (s/v_1-\Delta ) \\ {v}_1\cdot s=2\cdot v_1\cdot (s-\Delta \cdot v_1) \\ s=2(s-\Delta \cdot v_1) \\ 18=2(18-0.6\cdot v_1) \\ 18=2\cdot (18-0.6\cdot v_1) \\ 1.2v_1=18 \\ {v}_1=\frac{18}{1.2}=15 \\ {v}_1=15=15\text{km/h} \end{gathered}$$

$$\begin{gathered} v_2=2\cdot v_1=2\cdot 15=30=30\text{km/h} \\ \text{ Verifying Solution: } \\ {T}_1=s/v_1=18/15=\frac{6}{5}=1.2\text{h} \\ {T}_2=T_1-\Delta =1.2-0.6=\frac{3}{5}=0.6\text{h} \\ {s}_1=v_1\cdot T_1=15\cdot 1.2=18\text{km} \\ {s}_2=v_2\cdot T_2=30\cdot 0.6=18\text{km} \\ {s}_1=s_2=s \end{gathered}$$

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