The figure shows the $(x, t)$ and $(y, t)$ diagrams of a particle moving in $2$ dimensions. If the particle has a mass of $500 \, g$,find the force (magnitude and direction) acting on the particle.

(D) From figure $(a)$,the $(x, t)$ graph is a straight line passing through the origin,indicating constant velocity in the $x$-direction.
$v_x = \frac{\Delta x}{\Delta t} = \frac{2 - 0}{2 - 0} = 1 \, m/s$.
Since the velocity is constant,the acceleration in the $x$-direction is $a_x = 0$.
From figure $(b)$,the $(y, t)$ graph is a parabola passing through the origin,which follows the relation $y = kt^2$. At $t = 2 \, s$,$y = 4 \, m$,so $4 = k(2)^2$,which gives $k = 1$. Thus,$y = t^2$.
The velocity in the $y$-direction is $v_y = \frac{dy}{dt} = 2t$.
The acceleration in the $y$-direction is $a_y = \frac{dv_y}{dt} = 2 \, m/s^2$.
The mass of the particle is $m = 500 \, g = 0.5 \, kg$.
The force components are:
$F_x = m a_x = 0.5 \times 0 = 0 \, N$.
$F_y = m a_y = 0.5 \times 2 = 1 \, N$.
The magnitude of the resultant force is $F = \sqrt{F_x^2 + F_y^2} = \sqrt{0^2 + 1^2} = 1 \, N$.
The direction of the force is along the positive $y$-axis.