Recall that average velocity is equal to change in position over a given time interval,
[tex]\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}[/tex]
so that the x-component of [tex]\vec v_{\rm ave}[/tex] is
[tex]\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}[/tex]
and its y-component is
[tex]\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}[/tex]
Solve for [tex]x_2[/tex] and [tex]y_2[/tex], which are the x- and y-components of the copter's position vector after t = 1.60 s.
[tex]x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}[/tex]
[tex]y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}[/tex]
Note that I'm reading the given details as
[tex]x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}[/tex]
so if any of these are incorrect, you should make the appropriate adjustments to the work above.
