#### Answer

The car travels through an angle of 0.87 radians.

#### Work Step by Step

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 $a_c = 1.73~m/s^2$
$a_c = \frac{v^2}{r}$
$v = \sqrt{a_c~r}$
$v = \sqrt{(1.73~m/s^2)(120~m)}$
$v = 14.4~m/s$
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.