(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

I'm using:

d=1/2(Vf + Vi) * T

and:

v^2 = u^2 + 2as

3. 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!

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# Homework Help: Stopping distance given Vinitial and distance

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