Answer the question briefly

Dear Student,

Please find below the solution to the asked query:

Given that the acceleration of the particle changes with time as,

$a=3t^2+2t+2$
So, velocity varies with time as,

$$\begin{gathered} V=V_0+\int a dt = \int \left(3t^2+2t+2\right)dt \\ \Rightarrow V=V_0+t^3+t^2+2t \\ at t=0; V=2 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ \therefore V=t^3+t^2+2t+2 \end{gathered}$$

So, at t = 2 sec.

$$\begin{gathered} V\left(t\right)=t^3+t^2+2t+2 \Rightarrow V\left(t=2\right)=2^3+2^2+2\times 2+2=8+4+4+2 \\ \Rightarrow V\left(t=2\right)=18 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Position of particle at any time is,

$$\begin{gathered} X=X_0+\int V\left(t\right) dt = 2+{\int }_0^2\left(t^3+t^2+2t+2\right)dt \\ \Rightarrow X=2+{\left(\frac{t^4}{4}+\frac{t^3}{3}+t^2+2t\right)}_0^2=2+\left(\frac{16}{4}+\frac{8}{3}+4+4\right)=2+4+8+\frac{8}{3} \\ \Rightarrow X=\frac{50}{3} m \end{gathered}$$

Hope this information will clear your doubts about the topic.

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