# Stopping distance

1. Mar 29, 2012

### TalibanNinja

1. The problem statement, all variables and given/known data
A driver carelessly ignores the reduced speed limit of 40.0 km/h in a school zone and continues at 65 km/h. Assuming a good reaction time of 0.80 s, how many more metres will it take him to stop than if he had reduced his speed? Assume 2000kg car has a constant braking friction of 12000N. (Must use work-energy theorem)
va=11.11m/s
vb=18.05ms
Ff=12000N
m=2000kg
g=9.8m/s^2

2. Relevant equations

Ff=μFn
Work done to an object = change in kinetic energy
Wf=Ek
μmgd=1/2mv^2
d=v^2/2μg

3. The attempt at a solution

μ=Ff/Fn
μ=12000N/(2000kg*9.8m/s^2)
μ=0.61

da=(11.11m/s)^2/2*(0.61)(9.8m/s^2)
da=10.32m

db=(18.05m/s)^2/2*(0.61)(9.8m.s^2)
db=27.25m

Δd=db-da
Δd=27.25m - 10.32m
Δd=16.93

Therefore it takes the driver 16.93m longer to stop than if he slowed his speed.

Last edited: Mar 29, 2012
2. Mar 29, 2012

### tal444

You must show us an attempt at a solution first before we are allowed to help you. First try to write an equation for distance if he had slowed down to 40km/h.

3. Mar 29, 2012

### TalibanNinja

i've tried the question using the knowledge i have but im not sure.

4. Mar 29, 2012

### Staff: Mentor

Start by stating the work-energy theorem.

5. Mar 29, 2012

### tal444

Hint: Use $\frac{1}{2}$mv2=Fd