Space Shuttle and the Moon

Relevant wiki: Average Velocity

Using the very simple equation $time=\frac { distance }{ rate }$ , this problem can be tackled down very quickly. Plugging in what we have, we get $time=\frac { 2500\quad m/s }{ 3.6\times 10^{ 8 }m }$ .

Now, just convert the numerator into meters per day so the units left in our equation cancel out to days, we can approach this by perhaps dimensional analysis:

$time=\frac { 3.6\times 10^{ 8 }m }{ 2500\quad m/s } =\frac { 3.6\times 10^{ 8 }m }{ \frac { 2500m }{ 1s } \times \frac { 60^{ 2 }s }{ 1hr } \times \frac { 24hr }{ 1day } } =5/3\quad days$