Answer
The angle of the direction of travel relative to the positive direction of the x axis is $~~-125^{\circ}$
Work Step by Step
We can find the horizontal component of velocity:
$x(t) = -15.00+2.00t-4.00t^3$
$v_x(t) = \frac{dx}{dt} = 2.00-12.0t^2$
$v_x(0.700~s) = 2.00-(12.0)(0.700)^2 = -3.88~m/s$
We can find the vertical component of velocity:
$y(t) = 25.00+7.00t-9.00t^2$
$v_y(t) = \frac{dy}{dt} = 7.00-18.0~t$
$v_y(0.700~s) = (7.00)-(18.0)~(0.700) = -5.6~m/s$
We can find the direction of velocity as an angle below the negative direction of the x axis:
$tan~\theta = \frac{5.6}{3.88}$
$\theta = tan^{-1}~(\frac{5.6}{3.88})$
$\theta = 55.3^{\circ}$
Note that $~~180^{\circ} - 55.3^{\circ} \approx 125^{\circ}$
The angle of the direction of travel relative to the positive direction of the x axis is $~~-125^{\circ}$
