Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find the centripetal acceleration when the total acceleration is $2.0~m/s^2$ $a = \sqrt{(a_c)^2+(a_t)^2}$ $a^2 = (a_c)^2+(a_t)^2$ $a_c = \sqrt{a^2-(a_t)^2}$ $a_c = \sqrt{(2.0~m/s^2)^2-(1.0~m/s^2)^2}$ $a_c = 1.73~m/s^2$ We can find the speed when $a_c = 1.73~m/s^2$ as: $a_c = \frac{v^2}{r}$ $v = \sqrt{a_c~r}$ $v = \sqrt{(1.73~m/s^2)(120~m)}$ $v = 14.4~m/s$ We can find the distance around the curve that the car travels: $d = \frac{v^2-v_0^2}{2a}$ $d = \frac{(14.4~m/s)^2-0}{(2)(1.0~m/s^2)}$ $d = 104~m$ We can find the angle that the car travels: $\theta = \frac{d}{r} = \frac{104~m}{120~m}$ $\theta = 0.87~rad$ The car travels through an angle of 0.87 radians.