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 10:24. Find how far they will meet from A and when they will reach the destination.

Result

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

Step-by-step explanation:

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

3.4 · (v+20) + v·2.4 = 300

5.8v = 232


v = 232/5.8 = 40

v = 40

Our simple equation calculator calculates it.

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

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

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