Answer
At a distance of $~~7.5~m~~$ before the first point, the car was at rest.
Work Step by Step
We can find the distance the car moved as it accelerated from rest to a speed of $5.0~m/s$:
$v^2 = v_0^2+2ax$
$v^2 = 0+2ax$
$x = \frac{v^2}{2a}$
$x = \frac{(5.0~m/s)^2}{(2)(1.667~m/s^2)}$
$x = 7.5~m$
At a distance of $~~7.5~m~~$ before the first point, the car was at rest.
