Problem B and Solution

Into the Wild Blue Yonder!

Canadian Actor William Shatner travelled on the Blue Origin rocket in October \(2021\). He was in the rocket for \(10\) minutes and \(17\) seconds after liftoff, before landing back on the desert floor in Texas. The rocket rose to an altitude of \(105.9\) km.

If his flight was straight up and down, what was his mean speed, to the nearest kilometre per hour, over the course of the whole journey?

The length of the Trans-Canada Highway between the east and west coasts of Canada is \(7821\) km. If the rocket travels a distance of \(7821\) km at the mean speed found in part (a), approximately how long (in hours and minutes) would that trip take?

The total distance William Shatner travelled was \(105.9 \times 2=211.8\,\)km.

His travel time was \(10\) minutes and \(17\) seconds. Since there are \(60\) seconds in one minute, his travel time was \(10\times 60 +17=617\) seconds. Since there are \(60\times 60=3600\) seconds in each hour, his travel time in hours was \(617\div 3600\approx 0.1714\) hr.

Thus, his mean speed was \(211.8\mbox{ km}\div 0.1714\mbox{ hr}\approx 1236\) km/hr.

Travelling a distance of \(7821\) km at a mean speed of \(1236\) km/hr would take the rocket \(7821\div 1236\approx 6.328\) hr. Since there are \(60\) minutes in each hour, this is equal to \(6.328\times 60 \approx 380\) minutes, or approximately \(6\) hours and \(20\) minutes.

Note: Calculations here were carried out with four significant digits. Answers may vary if fewer are used at each stage.