## University Physics with Modern Physics (14th Edition)

(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.
(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.