The velocity a particle moving along the x-axis varies as a function of time t as $\mathrm{v}\left(\mathrm{t}\right) = (1– 3{\mathrm{t}}^2 + 2{\mathrm{t}}^3)\mathrm{m}/\mathrm{s}$. If its position at t = 0 is x = 0 then at t = 2s, its position is

Given $\mathrm{v}\left(\mathrm{t}\right) = (1– 3{\mathrm{t}}^2 + 2{\mathrm{t}}^3)\mathrm{m}/\mathrm{s}$

v = dx/dt

$\Rightarrow x\left(t\right) = \int vdt$

$\Rightarrow x\left(t\right) = \int (1-3t^2+2t^3)dt$

$\Rightarrow x\left(t\right) = t-t^3+\frac{t^4}{2}$   since x = 0 at t = 0, there will be no constant.

Therefore at t = 2secs, position is $2 - 2^3+\frac{2^4}{2}$= 2