A particle of mass $m={10}^{-3}\mathrm{kg}\text{ and charge }q=1\mathrm{C}$ is initially at rest. At time $t=0$, the particle starts experiencing an electric field $\overrightarrow{\mathrm{E}}=E_0\sin \left(\omega t\right)\hat{\mathit{i}}$  where $E_0=1\mathrm{N}/\mathrm{C}\text{ and }\omega ={10}^3\mathrm{rad}/\mathrm{s}$. The maximum speed (in m/s) attained by the particle is   

The instantaneous acceleration of the particle

is $a\left(t\right)=\frac{q{E}_0}{m}\sin \left(\omega t\right)$.  The instantaneous velocity is

given by 

$\begin{array}{l}v\left(t\right)=v\left(0\right)+{\int }_0^t a\left(t\right)dt\\ \Rightarrow v\left(t\right)=\frac{q{E}_0}{m}{\int }_0^t \sin \left(\omega t\right)dt\left[v\right(0)=0\text{, given }]\\ \Rightarrow v\left(t\right)=\frac{q{E}_0}{m\omega }[1-\cos (\omega t\left)\right]\end{array}$

The greatest speed is

$v_{max}=\frac{2q{E}_0}{m\omega }=\frac{2\left(1\mathrm{C}\right)(1\mathrm{N}/\mathrm{C})}{\left({10}^{-3}\mathrm{kg}\right)\left({10}^3\mathrm{rad}/\mathrm{s}\right)}=2 \mathrm{m}/\mathrm{s}$