An automobile is travelling on a long straight steady highway...?

An automobile is travelling on a long straight steady highway at 75.0mi/h when the driver sees a wreck 150m ahead. At that instant she applies the brakes (ignore reaction time). First there is 100m of ice where deceleration is only 1/00m/s^2. From then on she is on dry concrete where deceleration is 7.00m/s^2.
a) What is her speed just after leaving the ice portion of the road?
b) How much total distance does it take her to stop??
c) What is the total time it took her to stop??

I tried working on this problem for a bit but I am pretty stuck on it. I converted 75.0mi/h to m/s ---> I got 33.52m/s.
For A) I used the formula v^2=vo^2+2ad where I put v=? vo=33.52m/s and a =-1.00m/s (I think it is negative since it is decelerating). I got 30.39m/s for v.

for B and C I am totally stuck. I tried to reason out B but can't. I don't think I can find C without B. Any help would be awesome and I will surely pick the best answer shortly after!
Thanks for the answers guys!! I am looking at them now. Actually I had originally gotten 166m as the distance but I found it confusing because the crash site is 150m ahead, so wouldn't that mean she crashed into the crash site??

for the ice portion

100=33.5*t-.5*1*t^2
solve for t

3.13 seconds

v=33.5-1*3.13

30.4 m/s

then
0=30.4-7*t
t=30.4/7

total distance to stop:
s=100+.5*30.4^2/7

166 m

total time is
3.13+30.4/7

7.47 s



____________________-

75.0 mi/h = 33.52800 m / s from google calculator

a)
On ice

v^2 = (33.528)^2 - 2(100) = 924.12
answer v = 30.4 m/s

On concrete

(30.4)^2 = 0 - 2(7)x
924.12/ 14 = x = 66 meters

b)
100 +66 = 166 meters

c)

Use x =(1/2)at^2 for the time part by solvinfor t and applying this to each part of the pavement

t = sqrt(2ax)

on ice

t(ice) = sqrt(200) = 14.1 s

on concrete

t(concrete) = sqrt(2x7x66) = 30.4 s

Total 44.5 s