Revision as of 03:05, 20 January 2024 by Admin (Created page with "'''Answer: A''' <math>{ }_{1} V_{x}=A_{x+1}-P_{x} \ddot{a}_{x+1}=1-d \ddot{a}_{x+1}-P_{x} \ddot{a}_{x+1}</math> <math <math display="block"> =1-\underbrace{\left(P_{x}+d\right)} \ddot{a}_{x+1}=1-\ddot{a}_{x+1} / \ddot{a}_{x} </math> <math>\Rightarrow \ddot{a}_{x}\left(1-V_{x}\right)=\ddot{a}_{x+1}</math> Since <math>\ddot{a}_{x}=1+v p_{x} \ddot{a}_{x+1}</math> substituting we get <math>\ddot{a}_{x}\left(1-{ }_{1} V_{x}\right)=\frac{\ddot{a}_{x}-1}{v p_{x}} \Right...")
Exercise
ABy Admin
Jan 20'24
Answer
Answer: A
[math]{ }_{1} V_{x}=A_{x+1}-P_{x} \ddot{a}_{x+1}=1-d \ddot{a}_{x+1}-P_{x} \ddot{a}_{x+1}[/math]
[[math]]
=1-\underbrace{\left(P_{x}+d\right)} \ddot{a}_{x+1}=1-\ddot{a}_{x+1} / \ddot{a}_{x}
[[/math]]
[math]\Rightarrow \ddot{a}_{x}\left(1-V_{x}\right)=\ddot{a}_{x+1}[/math]
Since [math]\ddot{a}_{x}=1+v p_{x} \ddot{a}_{x+1}[/math] substituting we get
[math]\ddot{a}_{x}\left(1-{ }_{1} V_{x}\right)=\frac{\ddot{a}_{x}-1}{v p_{x}} \Rightarrow \ddot{a}_{x}\left(1-{ }_{1} V_{x}\right) v p_{x}=\ddot{a}_{x}-1[/math]
Solving for [math]\ddot{a}_{x}[/math], we get [math]\ddot{a}_{x}=\frac{1}{1-\left(1-{ }_{1} V_{x}\right) v p_{x}}=\frac{1}{1-(1-0.012)\left(\frac{1}{1.04}\right)(1-0.009)}[/math]
[[math]]
=17.07942
[[/math]]
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