Answer
Please see the work below.
Work Step by Step
We know that the displacement in your direction is:
$x=vt+\frac{1}{2}(acos(35))t^2$
We plug in the known values to obtain:
$x=(6.5)(6.3)+\frac{1}{2}(0.48)cos(35)(6.3)^2$
$x=48.75m$
The displacement in the perpendicular direction is:
$y=(\frac{1}{2})(a \space sin(35))t^2$
We plug in the known values to obtain:
$y=(\frac{1}{2})(0.48)sin(35)(6.3)^2=5.464m$
The magnitude of the displacement is
$d=\sqrt{x^2+y^2}$
We plug in the known values to obtain:
$d=\sqrt{(48.75)^2+(5.464)^2}=49m$
The direction of your displacement is:
$\theta=tan^{-1}\frac{y}{x}$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{5.464}{48.75})=6.4^{\circ}$
