It looks like you're given
[tex]x(t) = -11\,\mathrm m + \left(1\dfrac{\rm m}{\rm s}\right) t - \left(5\dfrac{\rm m}{\mathrm s^3}\right) t^3[/tex]
[tex]y(t) = 19\,\mathrm m + \left(3\dfrac{\rm m}{\rm s}\right) t - \left(9\dfrac{\rm m}{\mathrm s^2}\right) t^2[/tex]
The particle's position vector at time [tex]t[/tex] is given by
[tex]\vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
Differentiate [tex]\vec r[/tex] twice to recover the velocity and acceleration vectors.
[tex]\vec a(t) = \dfrac{d\vec v(t)}{dt} = \dfrac{d^2\vec r}{dt^2} = x''(t)\,\vec\imath + y''(t)\,\vec\jmath \\\\ \implies \vec v(t) = \left(1\dfrac{\rm m}{\rm s} - \left(15\dfrac{\rm m}{\mathrm s^3}\right) t^2\right)\,\vec\imath + \left(3\dfrac{\rm m}{\rm s} - \left(18\dfrac{\rm m}{\mathrm s^2}\right) t\right) \,\vec\jmath \\\\ \implies \vec a(t) = -\left(30\dfrac{\rm m}{\mathrm s^3}\right)t \, \vec\imath - \left(18\dfrac{\rm m}{\mathrm s^2}\right) \,\vec\jmath[/tex]
At [tex]t=1.4\,\rm s[/tex], the particle has acceleration
[tex]\vec a(1.4\,\mathrm s) = \left(-42\,\vec\imath - 18\,\vec\jmath) \dfrac{\rm m}{\mathrm s^2}[/tex]
with magnitude
[tex]\|\vec a(1.4\,\mathrm s)\| = \sqrt{\left(-42\dfrac{\rm m}{\mathrm s^2}\right)^2 + \left(-18\dfrac{\rm m}{\mathrm s^2}\right)^2} \approx 45.695\dfrac{\rm m}{\mathrm s^2}[/tex]
and making an angle [tex]\theta[/tex] relative to the positive [tex]x[/tex]-axis such that
[tex]\tan(\theta) = \dfrac{-18}{-42} = \dfrac37[/tex]
Since both components of the acceleration vector have negative sign, the acceleration points into the third quadrant, so that we add a multiple of 180° after taking the inverse tangent of both sides, namely
[tex]\theta = \tan^{-1}\left(\dfrac37\right) - 180^\circ \approx -156.801^\circ[/tex]
Now, (a) the magnitude of the net force acting on the particle is, by Newton's second law,
[tex]F = (0.41\,\mathrm{kg})\|\vec a(1.4\,\mathrm s)\| \approx \boxed{18.735\,\mathrm N}[/tex]
and (b) makes the same angle as the acceleration vector, [tex]\theta \approx \boxed{-156.801^\circ}[/tex].
At this moment, its velocity vector is
[tex]\vec v(1.4\,\mathrm s) = \left(-28.4\vec\imath - 22.2\,\vec\jmath\right) \dfrac{\rm m}{\rm s}[/tex]
which (c) makes an angle [tex]\theta[/tex] such that
[tex]\tan(\theta) = \dfrac{-22.2}{-28.4} = \dfrac{111}{142}[/tex]
This vector also points in the third quadrant, so
[tex]\theta = \tan^{-1}\left(\dfrac{111}{142}\right) -180^\circ \approx \boxed{-142.986^\circ}[/tex]
