A 4 kg mass moves under the influence of a force \(\vec{F}=\left(4 t^3 \hat{i}-3 t \hat{j}\right) N\) where \(t\) is the time in second. If mass starts from origin at \(t=0\), the velocity and position after \(t=2 s\) will be :

\(\overrightarrow{ F }=4 t^3 \hat{ i }-3 t \hat{ j }\)
\(\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }= t ^3 \hat{ i }-\frac{3}{4} \hat{ tj }\)
\(a_x=t^3\)
\(\frac{ dv _{ x }}{ dt }= t ^3\)
\(\int_{ v _{ x }=0}^{ v _{ x _2}} d v_{ x }=\int_{ t =0}^{ t =2} t^3 d t\)
\(v _{ x _2}-0=\left[\frac{ t ^4}{4}\right]_0^2\)
\(v_{x_2}=4\)
\(a_y=\frac{-3}{4} t\)
\(\frac{ dv _{ y }}{ dt }=-\frac{3}{4}\)
\(\int_0^{v_{v_2}} d v_y=\int_0^2 \frac{-3}{4} t d t\)
\(v _{ y _2}=\frac{-3}{4}\left[\frac{ t ^2}{2}\right]_0^2\)
\(v_{y_2}=\frac{-3}{2}\)
\(\vec{v}_2=4 \hat{i}-\frac{3}{2} \hat{j}\)
\(v_x=\frac{t^4}{4}\)
\(\int_0^{x_2} d x=\int_0^2 \frac{t^4}{4} d t\)
\(x_2-0=\left[\frac{t^5}{20}\right]_0^2\)
\(x_2=\frac{8}{5}\)
\(v_y=\frac{-3}{8} t^2\)
\(\int_0^{y_2} d x=\frac{-3}{8} t^2 d t\)
\(y _2-0=\frac{-3}{8}\left[\frac{ t ^3}{3}\right]_0^2\)
\(y_2=-1\)
\(\overrightarrow{ r }=\frac{8}{5} \hat{ i }-\hat{ j }\)