How fast does the vehicle go? (Almost worked version!)

This problem's question: How fast is the vehicle in knots rounded to the nearest integer?

There is a related problem .

The purpose of this problem is learning to deal with data in unexpected forms.

Fuel efficiencies are either volume per distance or distance per volume. The first case is effectively an area; the second case is effectively a per area. Either form can be converted to the other form. Here are usual fuel efficiencies converted to SI standard metric units:

${\color{#D61F06}\text{These conversions can be researched in Wolfram/Alpha.}}$ E.g, "1 liter per 100 kilometers in SI units" gives, among other answers: $1\times 10^{-8} m^2$ ".

$\frac{1\,\text{L}}{100\,\text{km}}=\frac{m^2}{10^8}$

$\frac{1\,\text{US gallon}}{1\,\text{US mile}}=\frac{24\times 10^{15}}{56451387097\,m^{-2}}$

The above converted to the metric form is about $\text{2.35214112904167E-6}\, m^2$ .

A knot is a unit more commonly used with ships. It is a unit of speed. In metric form, in SI units, it is $\frac{463}{900}\frac{m}{s}$ . In common metric form, it is $1852 \frac{m}{h}$ .

Fuel efficiency is 1 zettabarn. Yes, a barn is a real area unit. In fact, it is metric and common in high energy physics. A barn is $(10\,\text{fm})^2$ or $1\times 10^{-28}\,m^2$ . The zetta prefix makes a zettabarn be $\frac{1}{10}\,\text{mm}^2$ or $1\times 10^{-7}\,m^2$ .

Amount of fuel used is 1 oil barrel. Yes, that unit is used in the petroleum trade. If you use the metric equivalent, then the answer comes out the same when rounded to integer knots. One barrel of oil in SI units is commonly rounded to $0.159\,m^3$ or $159\,L$ . The exact metric definition is 0.158987 cubic meters. The petroleum trade definition is 42 US gallons, which converts to the exact metric definition.

Using the fuel efficiency, the distance traveled can be computed $\frac{0.158987\,m^3}{1\times 10^{-7}\,m^2}=1589870\,m$ or $1589.87\,\text{km}$ .

The trip took 17.5 hours exactly. The problem is only concerned with the average speed of the vehicle.

Because that we are only interested in the average speed, this is a distance-speed-time problem: $\text{speed}=\frac{\text{distance}}{\text{time}}$ .

$\frac{1589.87\,\text{km}}{17.5\,\text{h}}=90.8497142857143\,\frac{km}{h}$

We need the speed in knots, therefore the speed above needs to be divided by 1.852 to get knots.

Now, finish the problem!