Stopping distance given Vinitial and distance

[Staerke]

Homework Statement

It takes a minimum distance of 41.14 m to stop a car moving at 11.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Homework Equations

I'm using:
d=1/2(Vf + Vi) * T

and:

v^2 = u^2 + 2as

The Attempt at a Solution

So here we go, I'm going to calculate the time to stop using the first formula:
41.14 = 1/2(0+11) * T
T = 7.48

Cool, makes sense, alright. Now I'm going to use that number to get the acceleration:

0^2 = 11^2 + (2 * a * 7.48)
a = -8.088
Makes plenty of logical sense, am I right? Sweet,
Now I'm taking the acceleration and figuring out the stopping time of the faster moving vehicle like so

0^2 = 25^2+(2*-8.088*t)
t=38.64
This is weird, it takes 38 seconds for the car to stop, oh well let's see the distance

d = 1/2(Vf + Vi) * t
d = 1/2(0+25) * 38.64
d = 483 meters

Yeah no way...plug it in as the answer and no dice.

What am I doing wrong? Am I completely off here? Help!

 

[rl.bhat]

0^2 = 11^2 + (2 * a * 7.48)
a = -8.088
This step is wrong.
It should be
v^2 = u^2 -2*a*s, and s = 41.14 m

 

[mgb_phys]

You have mixed up the equations near the end
it's v^2 = u^2 + 2 a S , you seem to have done '2 a T'

As a rough estimate, the ke of the car is proprtional to v^2 so double the speed is 4x the energy = 4x the stopping distance

 

[Staerke]

0^2 = 11^2 + (2 * a * 7.48)
a = -8.088
This step is wrong.
It should be
v^2 = u^2 -2*a*s, and s = 41.14 m

I'm an idiot! Thanks for the help!