Answer:
The total duration of the trip is 48 hours.
Step-by-step explanation:
Let suppose that ship travels at constant speed during its travel. Each stage is represented by the following kinematic equation:
[tex]v =\frac{\Delta s}{\Delta t}[/tex]
Where:
[tex]\Delta s[/tex] - Travelled distance, measured in kilometers.
[tex]\Delta t[/tex] - Time, measured in hours.
[tex]v[/tex] - Speed, measured in kilometers per hour.
Now, each stage is represented by the following expressions:
Outbound trip
[tex]v = \frac{700\,km}{\Delta t}[/tex]
Return trip
[tex]v + 10\,\frac{km}{h} = \frac{700\,kh}{\Delta t - 8\,h}[/tex]
By eliminating [tex]v[/tex] and simplifying the resulting expression algebraically:
[tex]\frac{700\,km}{\Delta t} + 10\,\frac{km}{h} = \frac{700\,km}{\Delta t -8\,h}[/tex]
[tex](700\,km)\cdot \left(\frac{1}{\Delta t - 8\,h}-\frac{1}{\Delta t} \right) = 10\,\frac{km}{h}[/tex]
[tex]\frac{1}{\Delta t - 8\,h}-\frac{1}{\Delta t} = \frac{1}{70}\,\frac{1}{h}[/tex]
[tex]\frac{8\,h}{\Delta t \cdot (\Delta t-8\,h)} = \frac{1}{70}\,\frac{1}{h}[/tex]
[tex]560\,h^{2} = \Delta t\cdot (\Delta t - 8\,h)[/tex]
[tex](\Delta t )^{2}-8\cdot \Delta t - 560 = 0[/tex]
This equation can be solved by means of the Quadratic Formula, whose roots are presented below:
[tex]\Delta t_{1} = 28\,h[/tex] and [tex]\Delta t_{2} = -20\,h[/tex]
Only the first roots offers a physically resonable solution. Then, total duration of the trip is:
[tex]t_{T} = 28\,h +20\,h[/tex]
[tex]t_{T} = 48\,h[/tex]
The total duration of the trip is 48 hours.
