In a car, the brakes stop the tyres while friction between the tyres and the road surface stops the car. On a wet road the coefficient of friction between the road and the tyres is
0.1. two similar cars, A and B, are travelling at speeds of 15 m/s and 30 m/s
,respectively. Brakes are suddenly applied on each of the cars. How far will each of the cars travel before coming to a stop?
3 answers
plz I need the answer now
F = m a = - mu m g
so
a = - mu g = -9.81 mu
v = Vi - 9.81 mu t
0 = Vi - 9.81 mu t at stop
so stop time t = Vi / (9.81 mu)
distance = average speed * time
If Vi = 30, average speed = 15
distance = 15 ( 30 / ( 9.81 mu) ) = 15 *30 / 0.981
If Vi = 15, average speed = 7.5
distance = 7.5 * 15 / 0.981
so
a = - mu g = -9.81 mu
v = Vi - 9.81 mu t
0 = Vi - 9.81 mu t at stop
so stop time t = Vi / (9.81 mu)
distance = average speed * time
If Vi = 30, average speed = 15
distance = 15 ( 30 / ( 9.81 mu) ) = 15 *30 / 0.981
If Vi = 15, average speed = 7.5
distance = 7.5 * 15 / 0.981
Note that work done to stop is proportional to Kinetic energy at start
(1/2) m v^2
so stopping requires four times as far for twice the initial speed if the force is the same
(1/2) m v^2
so stopping requires four times as far for twice the initial speed if the force is the same
2 answers