The particle has acceleration vector
[tex]\vec a=\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\jmath[/tex]
We're told that it starts off at the origin, so that its position vector at [tex]t=0[/tex] is
[tex]\vec r_0=\vec0[/tex]
and that it has an initial velocity of 12 m/s in the positive [tex]x[/tex] direction, or equivalently its initial velocity vector is
[tex]\vec v_0=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\,\vec\imath[/tex]
To find the velocity vector for the particle at time [tex]t[/tex], we integrate the acceleration vector:
[tex]\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau[/tex]
[tex]\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath[/tex]
[tex]\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath[/tex]
Then we integrate this to find the position vector at time [tex]t[/tex]:
[tex]\vec r=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau[/tex]
[tex]\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath[/tex]
[tex]\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath[/tex]
Solve for the time when the [tex]y[/tex] coordinate is 18 m:
[tex]18\,\mathrm m=\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=3.0\,\mathrm s[/tex]
At this point, the [tex]x[/tex] coordinate is
[tex]\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)(3.0\,\mathrm s)+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.0\,\mathrm s)^2=27\,\mathrm m[/tex]
so the answer is C.
