A car is traveling at 55.5 km/hr on flat highway. If the coefficient of friction between road and tires on a rainy day is 0.187, what is the minimum distance in which the car will stop?
What is the stopping distance when the surface is dry and the coefficient of friction is 0.600?
What is the stopping distance when the surface is dry and the coefficient of friction is 0.600?
-
55.5kph = (55.5/3.6) = 15.42m/sec. (shortcut).
(9.8 x 0.187) = braking force of 1.833N.
Acceleration = (f/m) = 1.833/1, = -1.833m/sec^2.
Distance to stop = (v^2/2a) = 64.86 metres.
(9.8 x 0.6) = braking force of 5.88N.
Acceleration = (f/m) = 5.88/1, = -5.88m/sec^2.
Distance to stop = (v^2/2a) = 20.22 metres.
(9.8 x 0.187) = braking force of 1.833N.
Acceleration = (f/m) = 1.833/1, = -1.833m/sec^2.
Distance to stop = (v^2/2a) = 64.86 metres.
(9.8 x 0.6) = braking force of 5.88N.
Acceleration = (f/m) = 5.88/1, = -5.88m/sec^2.
Distance to stop = (v^2/2a) = 20.22 metres.