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- Thread starter chuck
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If we use the relationship between distance, constant speed, and time:

\(\displaystyle d=rt\)

and plug in the given data:

\(\displaystyle r=\frac{1.3\text{ ft}}{10\text{ s}}\)

\(\displaystyle t=30\text{ s}\)

we get:

\(\displaystyle d=\left(\frac{1.3\text{ ft}}{10\text{ s}} \right)\left(30\text{ s} \right)=3.9\text{ ft}\)