Answer
The stopping distance is 50.7 m.
She will not be able to stop in time. She will be at a distance of 30.7 m past the beginning of the intersection.
Work Step by Step
For the first 0.350 s, the acceleration is 0 because this is the driver's reaction time. We can calculate the distance traveled in the first 0.350 s:
$x = v_0t = (18.0 ~m/s)(0.350 ~s) = 6.30 ~m$
We can use $x_0 = 6.30 ~m$ in the next part of the question.
$x = x_0 + \frac{v^2-v_0^2}{2a} = 6.30 ~m + \frac{0-(18.0 ~m/s)^2}{(2)(-3.65 ~m/s^2)} = 50.7 ~m$
The stopping distance is 50.7 m. Since the driver is 20.0 m from the intersection, she will not be able to stop in time. She will be at a distance of 30.7 m past the beginning of the intersection.
