The particle has [tex]\vec r(0)=(4\,\mathrm m)\,\vec\imath[/tex] and [tex]\vec v(0)=\left(2\frac{\rm m}{\rm s}\right)\,\vec\imath[/tex], and is undergoing a constant acceleration of [tex]\vec a=-\left(2\frac{\rm m}{\mathrm s^2}\right)(\vec\imath+\vec\jmath)[/tex].
This means its position at time [tex]t[/tex] is given by the vector function,
[tex]\vec r(t)=\vec r(0)+\vec v(0)t+\dfrac12\vec a t^2[/tex]
[tex]\implies\vec r(t)=\left[4\,\mathrm m+\left(2\dfrac{\rm m}{\rm s}\right)t-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath[/tex]
The particle crosses the x-axis when the [tex]\vec\imath[/tex] component is 0 for some time [tex]t>0[/tex], so we solve:
[tex]4+2t-t^2=0\implies t^2-2t+1=5[/tex]
[tex]\implies(t-1)^2=5[/tex]
[tex]\implies t-1=\pm\sqrt5[/tex]
[tex]\implies t=1\pm\sqrt5[/tex]
The negative square root introduces a negative solution that we throw out, leaving us with [tex]\boxed{t=1+\sqrt5}[/tex] or about 3.24 seconds after it starts moving.
