Pedestrian cyclist meeting

Cities A and B are 36 km apart. A pedestrian leaves town A at a speed of 6 km/h to place B. A cyclist leaves town B simultaneously at a speed of 24 km/h to town A. In how many hours and at what distance will they meet? Write the result in minutes.

Final Answer:

t =  1.2 h
m =  72 min
s₁ =  7.2 km
s₂ =  28.8 km

Step-by-step explanation:

$$\begin{gathered} v_1=6\text{km/h} \\ {v}_2=24\text{km/h} \\ s=36\text{km} \\ v=v_1+v_2=6+24=30\text{km/h} \\ t=s/v=36/30=1.2\text{h} \end{gathered}$$

$m=t\to \text{min}=t\cdot 60\text{min}=1.2\cdot 60\text{min}=72\text{min}=72\text{min}$

$s_1=v_1\cdot t=6\cdot 1.2=7.2\text{km}$

$s_2=v_2\cdot t=24\cdot 1.2=28.8\text{km}$

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