SOLUTION: A car traveling at 60 mph requires 200 ft to stop. So if it is traveling at 100 mph how many ft does it take to stop. I have tried cross multiplication and keep coming to 333.33 ft

Question 146926: A car traveling at 60 mph requires 200 ft to stop. So if it is traveling at 100 mph how many ft does it take to stop. I have tried cross multiplication and keep coming to 333.33 ft. The answer I have for the problem is 556 ft. Please help. Thanks!
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Cross multiplying doesn't work because it's not a linear relationship.
Braking Distance on level pavement is given by the formula:
d=v^2/2gf where:
d=braking distance
v=speed of the car
g=acceleration due to gravity (32.2 ft/sec^2)
f=coefficient of friction between tires and road
Using the data that's given, we have:
200=60^2/64.4f multiply each side by 64.4f
12,880f=3600 divide each side by 12,880
f=0.2795----coefficient of friction between tires and road
Using the formula again, we have:
d=100^2/(64.4*0.2795)=
d=10,000/17.9998=555.562 ft
Hope this helps----ptaylor

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