# 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:

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

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