Revision as of 21:27, 19 January 2024 by Admin (Created page with "'''Answer: B''' Per equivalence Principle: <math display="block"> \begin{aligned} G \ddot{a}_{35} & =100,000 A_{35}+0.4 G+150+0.1 G \ddot{a}_{35}+50 \ddot{a}_{35} \\ 1770 \ddot{a}_{35} & =100,000\left(1-d \ddot{a}_{35}\right)+0.4(1770)+150+0.1(1770) \ddot{a}_{35}+50 \ddot{a}_{35} \\ 1770 \ddot{a}_{35} & =100,000+708+150+\ddot{a}_{35}\left(177+50-100,000\left(\frac{0.035}{1.035}\right)\right) \end{aligned} </math> Solving for <math>\ddot{a}_{35}</math>, we have <ma...")
Exercise
ABy Admin
Jan 19'24
Answer
Answer: B
Per equivalence Principle:
[[math]]
\begin{aligned}
G \ddot{a}_{35} & =100,000 A_{35}+0.4 G+150+0.1 G \ddot{a}_{35}+50 \ddot{a}_{35} \\
1770 \ddot{a}_{35} & =100,000\left(1-d \ddot{a}_{35}\right)+0.4(1770)+150+0.1(1770) \ddot{a}_{35}+50 \ddot{a}_{35} \\
1770 \ddot{a}_{35} & =100,000+708+150+\ddot{a}_{35}\left(177+50-100,000\left(\frac{0.035}{1.035}\right)\right)
\end{aligned}
[[/math]]
Solving for [math]\ddot{a}_{35}[/math], we have
[[math]]
\ddot{a}=\frac{100,858}{1770+3154.64}=\frac{100,858}{4924.64}=20.48
[[/math]]