The cruising speed of a 747 airplane is approximately 9,944 inches per second. What is this speed to the nearest foot per second?
For this case, the first thing we must do is take into account the following conversion factor:
\(1 \ ft = 12\ inches\)
Therefore, by applying the conversion factor for the given velocity we have:
\((9944 \frac {in} {s}) (\frac {1} {12} \frac {ft} {in}) = 829 \frac {ft} {s}\)
Answer:
The speed of the plane in foot per second is given by:
\(829 \frac {ft} {s}\)