Catching up time

Honza left home at 8 a.m. on foot at an average speed of 4 km/h. At 11 a.m., dad followed him on a bicycle at an average speed of 40 km/h. How long and at what distance from home will dad catch up with Honza?

Final Answer:

t =  0.3333 h
s =  13.3333 km

Step-by-step explanation:

$$\begin{gathered} v_1=4\text{km/h} \\ {v}_2=40\text{km/h} \\ \Delta t=11-8=3\text{h} \\ {s}_1=s_2 \\ {v}_1(t+\Delta t)=v_2\cdot t \\ 4(t+3)=40\cdot t \\ 4\cdot (t+3)=40\cdot t \\ 36t=12 \\ t=\frac{12}{36}=0.33333333 \\ t=\frac{1}{3}\doteq 0.333333=0.3333\text{h} \end{gathered}$$

$s=v_2\cdot t=40\cdot 0.3333=13.3333\text{km}$

Try another example