Answer
The angle of the net force relative to the positive direction of the x axis is $~~-133^{\circ}$
Work Step by Step
We can find the horizontal component of acceleration:
$x(t) = -15.00+2.00t-4.00t^3$
$v_x(t) = \frac{dx}{dt} = 2.00-12.0t^2$
$a_x(t) = \frac{dv_x}{dt} = -24.0~t$
$a_x(0.700~s) = (-24.0)(0.700) = -16.8~m/s^2$
We can find the vertical component of acceleration:
$y(t) = 25.00+7.00t-9.00t^2$
$v_y(t) = \frac{dy}{dt} = 7.00-18.0~t$
$a_y(t) = \frac{dv_y}{dt} = -18.0$
$a_y(0.700~s) = -18.0~m/s^2$
We can find the direction of acceleration as an angle below the negative direction of the x axis:
$tan~\theta = \frac{18.0}{16.8}$
$\theta = tan^{-1}~(\frac{18.0}{16.8})$
$\theta = 47^{\circ}$
The angle relative to the positive direction of the x axis is $~~-133^{\circ}$
Since the direction of the net force on the particle is equal to the direction of the acceleration, the angle of the net force relative to the positive direction of the x axis is $~~-133^{\circ}$
