Answer:
[tex]V_{avg} = \frac{ln2}{e} = 0.255fts^{-1}[/tex]
Step-by-step explanation:
The question asks for the average velocity and not the instantaneous velocity (which would have meant to differentiate). So, the right formula is
[tex]V_{avg} = \frac{s(t_{2}) - s(t_{1} ) }{t_{2} -t_{1} } =[/tex]
From the question, [tex]t_{1}[/tex] corresponds to e seconds and [tex]t_{2}[/tex] corresponds to 2e seconds. So, we have
[tex]V_{avg} = \frac{ln(2e) - ln(e)}{2e - e}[/tex]
One of the laws of logarithms says that
[tex]ln(\frac{a}{b} )= ln(a) - ln(b). \\\\ Therefore, ln(2e) - ln(e) = ln(\frac{2e}{e} ) = ln(2)[/tex]
[tex]V_{avg} = \frac{ln(2)}{e} = 0.2550fts^{-1}[/tex] ≅ [tex]0.255fts^{-1}[/tex]
