# 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:

$s=300 \ \text{km} \ \\ s_{1} + s_{2}=s \ \\ s_{1}=(v+20)t \ \\ s_{2}=v(t-1) \ \\ \ \\ t=(10+24/60) - 7.00=\dfrac{ 17 }{ 5 }=3.4 \ \text{h} \ \\ \ \\ (v+20)t + v(t-1)=300 \ \\ \ \\ 3.4 \cdot \ (v+20) + v \cdot \ 2.4=300 \ \\ 5.8v=232 \ \\ \dfrac{ 29 }{ 5 }v=232 \ \\ 29v=1160 \ \\ v=1160 / 29=40 \ \\ \ \\ \ \\ s_{1}=(v+20) \cdot \ t=(40+20) \cdot \ 3.4=204 \ \text{km}$
$t_{1}=7.00 + s / (v+20)=7.00 + 300 / (40+20)=12:00 \ \text{hh}:\text{mm}$
$t_{2}=8.00 + s / v=8.00 + 300 / 40=\dfrac{ 31 }{ 2 }=15.5=15:30 \ \text{hh}:\text{mm}$

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