Time it takes for a car to stop?

Question

I don't understand these...PLEASE HELP!!! thanks

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35m/s. At the moment the cars are 45m apart, the lead driver applies the brakes, causing the car to have an acceleration of –2.0m/s(squared).
a) How long does it take for the lead car to stop?
b) Assume that the driver of the chasing car applies the brakes at the same time as the driver of the lead car. What must the chasing car’s minimum negative acceleration be to avoid hitting the lead car?
c) How long does it take the chasing car to stop?

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 13.0m/s(squared). At t1 the rocket engine is shut down and the sled moves with constant velocity v until t2. The total distance traveled by the sled is 5300 m and the total time is 90.0s. Find t1, t2, and v

Answer 1

Look at this problem. It's exactly the same as yours except the numbers are slightly different. Just change the numbers.

http://answers.yahoo.com/question/index;_ylt=Al_EtjKtW3hpxhe.6vrZSmLty6IX;_ylv=3?qid=20070911194721AAUa8U0&show=7#profile-info-Iuz6G1dEaa

Problem 2:
Let v be the maximum velocity achieved. Then:
Leg1 average velocity =[ v(zero) + v(final)]/2 = v/2
Leg2 time:
v = at = 13t1
t1 = v/13

Total distance:
D = v1t1 + v2t2
D = (v/2)(v/13) + v(90 - (v/13))
5300 = v^2/26 + 90v -v^2/13
-v^2/26 + 90v -5300 = 0
v^2 -2340v + 13700 = 0
Use the quadratic formula:
v = (-B +/- √[B^2 - 4AC])/2A
v = (2340 +/- √[2340^2 - 4(137800)])/2
v = (2340 +/- 2219.10)/2
v = 2279.55m/s ; 60.45m/s
The first answer is too big.
t1 = v/13 = 60.45/13 = 4.65sec
t2 = 90 - t1 = 90 - 4.65 = 85.35sec

Check:
D = v1t1 + v2t2 = (60.45/2)(4.65) + (85.35)(60.45) = 5300m