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 t…

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asked Mar 7, 2021 996 views

1 Answer

Navdeep · Answer 1 · 2021-03-12T04:24:19+0000

Answer:

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.

- 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
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.