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?
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
Well, let me calculate that for you, but I have to warn you, my calculations might get a bit slippery!
Assuming constant deceleration, we can find the distance traveled using the equation: distance = (initial velocity^2) / (2 * acceleration)
For car A:
Acceleration = (0 - 15 m/s) / (2 * 0.1) = -150 m/s^2
Distance = (15 m/s)^2 / (2 * -150 m/s^2) = -1.25 meters
For car B:
Acceleration = (0 - 30 m/s) / (2 * 0.1) = -300 m/s^2
Distance = (30 m/s)^2 / (2 * -300 m/s^2) = -3.75 meters
Now, remember that negative distance means the car will be moving backward, so we can just take the absolute value to get the actual distance traveled.
So, car A will travel approximately 1.25 meters before stopping, and car B will travel approximately 3.75 meters. Just keep in mind that these are idealized calculations, and in reality, many other factors could affect the stopping distance. Drive safely!
To determine how far each car will travel before coming to a stop, we need to use the distance formula:
distance = (speed^2) / (2 * deceleration)
First, let's calculate the deceleration for each car using the coefficient of friction and the gravitational acceleration (assumed to be 9.8 m/s^2):
deceleration = coefficient of friction * gravitational acceleration
For car A:
deceleration_A = 0.1 * 9.8 = 0.98 m/s^2
For car B:
deceleration_B = 0.1 * 9.8 = 0.98 m/s^2
Now, we can calculate the distance each car will travel:
For car A:
distance_A = (speed_A^2) / (2 * deceleration_A)
= (15^2) / (2 * 0.98)
= 17.35 meters
For car B:
distance_B = (speed_B^2) / (2 * deceleration_B)
= (30^2) / (2 * 0.98)
= 183.67 meters
Therefore, car A will travel approximately 17.35 meters before coming to a stop, and car B will travel approximately 183.67 meters before coming to a stop.
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