Alonzo Odom

Answered

2022-07-26

$\mathrm{\u25b3}X={V}_{0}T+1/2a{T}^{2}$

$-45m=10m/sT-4.90m/{s}^{2}{T}^{2}$

$-45m=10m/sT-4.90m/{s}^{2}{T}^{2}$

Answer & Explanation

umshikepl

Expert

2022-07-27Added 11 answers

Since you have two equations, and both have a T and ${T}^{2}$ term, you can rewrite the second equation to have ${T}^{2}$ interms of $T:-4.90m/{s}^{2}{T}^{2}=-45m-10m/sT$.

Multiply through by $(-1):4.90m/{s}^{2}{T}^{2}=45m+10m/sT$.

Divide through by $(4.90m/{s}^{2}):{T}^{2}=(45{s}^{2}/4.90)+(10s/4.90)T$.

Now, plug this equation into the first equation: $\mathrm{\u25b3}X={V}_{0}T+1/2a{T}^{2}={V}_{0}T+1/2a[(45{s}^{2}/4.90)+(10s/4.90)T]$.

Multiply through: $\mathrm{\u25b3}X={V}_{0}T+45{s}^{2}/9.80+10sT/9.80$.

Multiply all by $(9.80):9.80\mathrm{\u25b3}X=9.80{V}_{0}T+45{s}^{2}+10sT$.

Add like terms: $9.80\mathrm{\u25b3}X=T(9.80{V}_{0}+10s)+45{s}^{2}$. Your solution, then, gives $\mathrm{\u25b3}X$ in terms of T and ${V}_{0}$.

Multiply through by $(-1):4.90m/{s}^{2}{T}^{2}=45m+10m/sT$.

Divide through by $(4.90m/{s}^{2}):{T}^{2}=(45{s}^{2}/4.90)+(10s/4.90)T$.

Now, plug this equation into the first equation: $\mathrm{\u25b3}X={V}_{0}T+1/2a{T}^{2}={V}_{0}T+1/2a[(45{s}^{2}/4.90)+(10s/4.90)T]$.

Multiply through: $\mathrm{\u25b3}X={V}_{0}T+45{s}^{2}/9.80+10sT/9.80$.

Multiply all by $(9.80):9.80\mathrm{\u25b3}X=9.80{V}_{0}T+45{s}^{2}+10sT$.

Add like terms: $9.80\mathrm{\u25b3}X=T(9.80{V}_{0}+10s)+45{s}^{2}$. Your solution, then, gives $\mathrm{\u25b3}X$ in terms of T and ${V}_{0}$.

Most Popular Questions