Answer:
\[{{s}_{n}}=u+\frac{a}{2}(2n-1)\]
\[20=u+\frac{a}{2}(2-1)\] \[(\because n=1s)\]
\[20=u\frac{a}{2}\] ………….. (1)
\[30=u+\frac{a}{2}(2\times 2-1)\] \[(\because n=2s)\]
\[30=u+\frac{3a}{2}\] ………… (2)
Solving (1) & (2) for ‘a’,
\[\frac{\begin{align}
& 30=u+\frac{3a}{2} \\
& 20=u+\frac{a}{2} \\
\end{align}}{10=a}\]
\[\because a=10m{{s}^{-2}}\]
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