Answer
(a) When t = 2.31 s, the velocity vector makes an angle of $30.0^{\circ}$ clockwise from the +x-axis.
(b) The acceleration vector has a magnitude of $0.275~m/s^2$ in the direction of $49.1^{\circ}$ clockwise from the +x-axis.
Work Step by Step
(a) $x(t) = (2.90~m)+ (0.0900~m/s^2)~t^2$
$v_x(t) = \frac{dx}{dt} = (0.1800~m/s^2)~t$
$y(t) = -(0.0150~m/s^3)~t^3$
$v_y(t) = \frac{dy}{dt} = -(0.0450~m/s^3)~t^2$
If the velocity vector makes an angle of $30.0^{\circ}$ clockwise from the +x-axis, then:
$\frac{v_y}{v_x} = -tan(30.0^{\circ})$
$\frac{-(0.0450~m/s^3)~t^2}{(0.1800~m/s^2)~t} = -tan(30.0^{\circ})$
$t = \frac{(0.1800~m/s^2)~tan(30.0^{\circ})}{(0.0450~m/s^3)}$
$t = 2.31~s$
When t = 2.31 s, the velocity vector makes an angle of $30.0^{\circ}$ clockwise from the +x-axis.
(b) $v_x(t) = \frac{dx}{dt} = (0.1800~m/s^2)~t$
$a_x(t) = \frac{dv_x}{dt} = 0.1800~m/s^2$
$v_y(t) = \frac{dy}{dt} = -(0.0450~m/s^3)~t^2$
$a_y(t) = \frac{dv_y}{dt} = -(0.0900~m/s^3)~t$
At t = 2.31 s:
$a_x(t) = 0.1800~m/s^2$
$a_y(t) = -(0.0900~m/s^3)(2.31~s) = -0.208~m/s^2$
We can find the magnitude of the acceleration vector.
$a = \sqrt{(0.1800~m/s^2)^2+(-0.208~m/s^2)^2}$
$a = 0.275~m/s^2$
We can find the angle clockwise from the +x-axis of the acceleration vector.
$tan(\theta) = \frac{0.208~m/s^2}{0.1800~m/s^2}$
$\theta = tan^{-1}(\frac{0.208}{0.1800}) = 49.1^{\circ}$
The acceleration vector has a magnitude of $0.275~m/s^2$ in the direction of $49.1^{\circ}$ clockwise from the +x-axis.
