Question

The driver of a 1730 kg car traveling on a horizontal road at 120 km/h suddenly applies the brakes. Due to a slippery pavement, the friction of the road on the tires of the car, which is what slows down the car, is 27.0 % of the weight of the car.(A)What is the acceleration of the car?(B)How many meters does it travel before stopping under these conditions?

Answer 1

Hey Welcome to Open Study :D

Answer 2

Is this math?

Answer 3

no physics

Answer 4

I am confused, what has 27 percent of the cars weight also the car itself will not have any acceleration because it is slowing down and acceleration is therefor define as an increase of ones speed over a certain amount of time. But as for B best why would be using this equation

d = V2/(2g(f + G))
Where:
d = Braking Distance (ft)
g = Acceleration due to gravity (32.2 ft/sec2)
G = Roadway grade as a percentage; for 2% use 0.02
V = Initial vehicle speed (ft/sec)
f = Coefficient of friction between the tires and the roadway

The table below gives a few values for the frictional coefficient under wet roadway surface conditions.

Design Speed (mph) Coefficient of Friction (f)
20 0.40
30 0.35
40 0.32
60 0.29

120 Kilometer/Hour (km/h) = 109.36133 Foot/Second (ft/s)

So plugin the equation d = 109.36*2/(2*32.2(f + .27))

I think you can fill in the rest but I hope it helped I spent a good amount of time to hopefully explain this to you.