Question

A ship travels a distance of 700 km. On the return trip it averages 10km/hr faster and 8 hours less, tp travel the 700km back. Determine how long the original part of the trip took in hours

Answer 1

The total duration of the trip is 48 hours.

Explanation:

Let suppose that ship travels at constant speed during its travel. Each stage is represented by the following kinematic equation:

`v =(\Delta s)/(\Delta t)`

Where:

`\Delta s` - Travelled distance, measured in kilometers.

`\Delta t` - Time, measured in hours.

`v` - Speed, measured in kilometers per hour.

Now, each stage is represented by the following expressions:

Outbound trip

`v = (700\,km)/(\Delta t)`

Return trip

`v + 10\,(km)/(h) = (700\,kh)/(\Delta t - 8\,h)`

By eliminating
`v` and simplifying the resulting expression algebraically:

`(700\,km)/(\Delta t) + 10\,(km)/(h) = (700\,km)/(\Delta t -8\,h)`

`(700\,km)\cdot \left((1)/(\Delta t - 8\,h)-(1)/(\Delta t) \right) = 10\,(km)/(h)`

`(1)/(\Delta t - 8\,h)-(1)/(\Delta t) = (1)/(70)\,(1)/(h)`

`(8\,h)/(\Delta t \cdot (\Delta t-8\,h)) = (1)/(70)\,(1)/(h)`

`560\,h^(2) = \Delta t\cdot (\Delta t - 8\,h)`

`(\Delta t )^(2)-8\cdot \Delta t - 560 = 0`

This equation can be solved by means of the Quadratic Formula, whose roots are presented below:

`\Delta t_(1) = 28\,h` and
`\Delta t_(2) = -20\,h`

Only the first roots offers a physically resonable solution. Then, total duration of the trip is:

`t_(T) = 28\,h +20\,h`

`t_(T) = 48\,h`

The total duration of the trip is 48 hours.