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 Wrong answer
t2 = 15:30 hh:mm Wrong answer

Solution:

s=300 km s1+s2=s s1=(v+20)t s2=v(t1)  t=(10+24/60)7.00=175=3.4 h  (v+20)t+v(t1)=300  3.4 (v+20)+v 2.4=300 5.8v=232 295v=232 29v=1160 v=1160/29=40   s1=(v+20) t=(40+20) 3.4=204=204  km 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 }
t1=7.00+s/(v+20)=7.00+300/(40+20)=12=12:00  hh:mm t_{ 1 } = 7.00 + s / (v+20) = 7.00 + 300 / (40+20) = 12 = 12:00 \ \text{ hh:mm }
t2=8.00+s/v=8.00+300/40=312=15.5=15:30  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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