#### Answer

The particle's velocity at x = 3 m is 11.1 m/s

#### Work Step by Step

The work done on the particle is equal to the area under the force versus position graph.
We can find the work done during the interval 0 - 3 m:
$W = (15~N)(1~m)+\frac{1}{2}(15~N)(2~m)$
$W = 30~J$
We can use the work-energy theorem to find the particle's velocity at x = 3 m.
$KE_2 = KE_1+W$
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W$
$v_2^2 = \frac{mv_1^2+2W}{m}$
$v_2 = \sqrt{\frac{mv_1^2+2W}{m}}$
$v_2 = \sqrt{\frac{(0.50~kg)(2.0~m/s)^2+(2)(30~J)}{0.50~kg}}$
$v_2 = 11.1~m/s$
The particle's velocity at x = 3 m is 11.1 m/s