The energy of a system as a function of time t is given as E (t) =...

Energy $E = {A}^{2}{e}^{- \alphat}$
For small % errors, we can, do differentiation
$dE = 2A\left(dA\right){e}^{- \alphat} + {A}^{2}$\left(- \alpha{e}^{- \alphat}dt\right)$<br >Fractional error$= \dfrac{dE}{E} = \dfrac{2A{e}^{- \alphat}\left(dA\right) + \left(- \alpha{A}^{2}{e}^{- \alphat}\right)dt}{{A}^{2}{e}^{- \alphat}} = 2\left(\dfrac{dA}{A}\right) + \left(- \alpha\dfrac{dt}{t}\right)t
% error = 2\left(1.25%\right) + \left(0.2 \times 1.5%\right) \times 5
$= 4$ (errors always add up)
Alternate solution:
$E = {A}^{2}{e}^{- \alphat}$
Taking natural logarithm on both sides,
$\ln \left(E\right) = ln$\left({A}^{2}\right) + \left(- \alphat\right)$<br >Differentiating<br >$\dfrac{dE}{E} = 2\left(\dfrac{dA}{A}\right) + $\left(- \alphadt\right)$
For small fractional erros, errors always add up
\left|\dfrac{dE}{E}\right| = 2$\left|\dfrac{dA}{A}\right| + \alpha\left(\dfrac{dt}{t}\right) \times t$
= 2\left(1.25%\right) + \left(0.2\right)\left(1.5%\right)5
$= 4$