W = ${\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}} = \frac{1}{2}\mathrm{m}({{\mathrm{v}}_{\mathrm{f}}}^2-{{\mathrm{v}}^2}_{\mathrm{i}})$

Area under acceleration-time graph is change in velocity.

$\begin{array}{rcl}{\mathrm{v}}_{\mathrm{f}}-0& =& \frac{1}{2}\mathrm{base}\times \mathrm{height}=\frac{1}{2}{\mathrm{t}}_2{\mathrm{a}}_2\\ {\mathrm{v}}_{\mathrm{i}}& =& \frac{1}{2}{\mathrm{t}}_1{\mathrm{a}}_1\end{array}$

So, $\mathrm{W} = \frac{1}{2}\mathrm{m}[{\left(\frac{1}{2}{\mathrm{a}}_2{\mathrm{t}}_2\right)}^2-{\left(\frac{1}{2}{\mathrm{a}}_1{\mathrm{t}}_1\right)}^2]$

             = $\frac{1}{8}\mathrm{m}[\frac{{{\mathrm{t}}^4}_2}{{{\mathrm{t}}^2}_1}{{\mathrm{a}}^2}_1-{{\mathrm{a}}^2}_1{{\mathrm{t}}^2}_{1 }] (\mathrm{since}, {\mathrm{a}}_{2 } = \frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}{\mathrm{a}}_1)$

             = $\frac{{{\mathrm{ma}}^2}_1}{8{{\mathrm{t}}^2}_1}({{\mathrm{t}}^4}_2-{{\mathrm{t}}^4}_1)$