Two ships

The distance from A to B is 300km. At 7 am started from A to B a ferry with speed higher by 20 km/h than a ship that leaves at 8 o'clock from B to A. Both met at 10h 24min. Determine how far they will meet from A and when they reach the destination.

Result

s₁ =  204 km
t₁ = 12:00 hh:mm
t₂ = 15:30 hh:mm

Solution:

$$\begin{gathered} s=300\text{ km } \\ {s}_1+s_2=s \\ {s}_1=(v+20)t \\ {s}_2=v(t-1) \\ t=(10+24\mathrm{/}60)-7.00={\displaystyle \frac{17}{5}}=3.4\text{ h } \\ (v+20)t+v(t-1)=300 \\ 3.4\cdot (v+20)+v\cdot 2.4=300 \\ 5.8v=232 \\ {\displaystyle \frac{29}{5}}v=232 \\ 29v=1160 \\ v=1160\mathrm{/}29=40 \\ {s}_1=(v+20)\cdot t=(40+20)\cdot 3.4=204\text{ km} \end{gathered}$$

$t_1=7.00+s\mathrm{/}(v+20)=7.00+300\mathrm{/}(40+20)=12=12:00\text{ hh}:\text{mm}$

$t_2=8.00+s\mathrm{/}v=8.00+300\mathrm{/}40={\displaystyle \frac{31}{2}}=15.5=15:30\text{ hh}:\text{mm}$

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