Alonzo Odom

2022-07-26

$\mathrm{△}X={V}_{0}T+1/2a{T}^{2}$
$-45m=10m/sT-4.90m/{s}^{2}{T}^{2}$

umshikepl

Expert

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 $\left(-1\right):4.90m/{s}^{2}{T}^{2}=45m+10m/sT$.
Divide through by $\left(4.90m/{s}^{2}\right):{T}^{2}=\left(45{s}^{2}/4.90\right)+\left(10s/4.90\right)T$.
Now, plug this equation into the first equation: $\mathrm{△}X={V}_{0}T+1/2a{T}^{2}={V}_{0}T+1/2a\left[\left(45{s}^{2}/4.90\right)+\left(10s/4.90\right)T\right]$.
Multiply through: $\mathrm{△}X={V}_{0}T+45{s}^{2}/9.80+10sT/9.80$.
Multiply all by $\left(9.80\right):9.80\mathrm{△}X=9.80{V}_{0}T+45{s}^{2}+10sT$.
Add like terms: $9.80\mathrm{△}X=T\left(9.80{V}_{0}+10s\right)+45{s}^{2}$. Your solution, then, gives $\mathrm{△}X$ in terms of T and ${V}_{0}$.

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