if a car is traveling at 30 miles per hour on a slippery road, then its stopping distanct D in feet can be caluclated by D=900/30*1/x where x is the coefficient of friction between the tires and road 01 multiply and simplfy the fromula for D compare the stopping distance on an icy road with x=0.1 and on dry pavement with x=0.8

Question

if a car is traveling at 30 miles per hour on a slippery road, then its stopping distanct D in feet can be caluclated by D=900/30*1/x where x is the coefficient of friction between the tires and road 0<x>1 multiply and simplfy the fromula for D
compare the stopping distance on an icy road with x=0.1 and on dry pavement with x=0.8

anonymous · Answer

Why is there a *1 in that equation?

radar · Answer

$$D=900/30^{1/x}$$ Is that the equation?

anonymous · Answer

yes

anonymous · Answer

the the 1/x is an exponent?.....

radar · Answer

$$D=30^{(2x-1)/x)}$$

radar · Answer

for x=.1
$$D=30^{-8}$$
for x=.8
$$D=30^{.75}$$
Doesn't seem right, thus am not positive about this!

radar · Answer

Someone else needs to pitch in here.

radar · Answer

$$900\over30^{1/x}$$
equals
$$30^{2}\over30^{1/x}$$
$$30^{(2-1/x)}$$

radar · Answer

The above was some of my reasoning (though it may be unreasonable lol)