Answer
${\theta_v}=28.0^{\circ}$
Work Step by Step
Given that,
$v=(8.00ti^{\wedge}+3.00t^2j^{\wedge})$
We plug in the value of $t=1.415$ to obtain:
$v=(8.00(1.415)i^{\wedge}+3.00(1.415)^2j^{\wedge})$
$v=(11.3i^{\wedge}+6.01j^{\wedge})$
Thus we obtain:
${\theta_v}=tan^{-1}\frac{v_y}{v_x}$
We plug in the known values to obtain:
${\theta_v}=tan^{-1}\frac{6.01}{11.3}$
${\theta_v}=28.0^{\circ}$
