Answer to Question #289376 in Mechanics | Relativity for ANDREW E. KIM

Question

8) On its maiden voyage in July 1952, the liner United States won the coveted Blue Ribbon for the fastest crossing of the Atlantic from New York to Cornwall, U.K. The trip took 3 days 10 hours 40 min at an average speed of 34.5 knots (65.5 km/h). This was 10 h 2 min less than the 14-year-old record held by the Queen Mary. What was the average speed of the Queen Mary

Answer 1

Explanation & Calculation

• The time it would have taken by Queen Mary is

$\qquad\qquad \begin{aligned} \small T_{QM}&=\small 3D10H40M+0D10H2M\\ &=\small 3D20H42M \end{aligned}$

• Before, going any further it is needed to express all the given times in hours to match the units of the speed ($\small \text{km/h}$ ).
• Then,

$\qquad\qquad \begin{aligned} \small T_{L-US}&=\small 3D10H40M\\ &=\small (3\times24h)+(10h)+(\frac{40}{60}h)\\ &=\small \frac{248}{3}h\\ \\ T_{QM}&=\small 3D10h42M\\ &=\small (3\times24h)+(10h)+(\frac{42}{60}h)\\ &=\small \frac{827}{10}h \end{aligned}$

• The length between New York and Cornwell can be calculated by the details given for the Liner US.

$\qquad\qquad \begin{aligned} \small d&=\small U_{L-US} \times T_{L-US}\\ &=\small 65.5\,\text{km/h}\times\frac{248}{3}h\\ \end{aligned}$

• Then the average speed of Queen Mary is

$\qquad\qquad \begin{aligned} \small V_{QM}&=\small \frac{\text{distance}}{\text{time}}\\ &=\small \frac{65.5\,km/h\times\frac{248}{3}h}{\frac{827}{10}h}\\ &=\small 65.47\,\text{km/h} \end{aligned}$