Answer
$r = (4.50~m)\hat{i}-(2.25~m)\hat{j}$
Work Step by Step
We can find the time when $v_x = 0$:
$v_{xf} = v_{i0}+a_x~t$
$t = \frac{v_{xf} - v_{i0}}{a_x}$
$t = \frac{0 - 3.00~m/s}{-1.00~m/s^2}$
$t = 3.00~s$
We can find $x$ at $t = 3.00~s$:
$x = v_{x0}~t+\frac{1}{2}a_x~t^2$
$x = (3.00~m/s)(3.00~s)+\frac{1}{2}(-1.00~m/s^2)(3.00~s)^2$
$x = 4.50~m$
We can find $y$ at $t = 3.00~s$:
$y = v_{y0}~t+\frac{1}{2}a_y~t^2$
$y = (0)(3.00~s)+\frac{1}{2}(-0.500~m/s^2)(3.00~s)^2$
$y = -2.25~m$
We can write the position vector in unit-vector notation:
$r = (4.50~m)\hat{i}-(2.25~m)\hat{j}$
