A car of mass 1550 kg is travelling at 75.0 km/h on a dry horizontal concrete surface when the driver applies the brakes licking the wheels. (a) How far does the car travel before coming to rest? (b) How long does it take the car to stop after the driver applies the brakes?
Coefficient of friction: static=0.948 kinetic=0.709
Why would he lick the wheels? Do you mean that he LOCKS the wheels?
(a) Initial kinetic energy = (Friction force)*(length of skid)
M*V^2/2 = (0.709)*M*g*X
M cancels out. Solve for X
X = V^2/[(1.418*g]
(b) Initial momentum = (Friction force)*(Time of skid)
M*V = (0.709)M*g*T
T = V/(0.709g)
m*155
t*55 u 36 155*55*36
To answer these questions, we need to use the concept of Newton's laws of motion, specifically the equations of motion for an object under constant acceleration. To find the distance traveled by the car before coming to a stop, we can use the equation:
s = (v^2 - u^2) / (2a)
Where:
s = distance traveled
v = final velocity (0 m/s when the car comes to rest)
u = initial velocity (75.0 km/h converted to m/s)
a = acceleration (negative, due to braking)
Step 1: Convert the initial velocity from km/h to m/s.
Using the conversion factor: 1 km/h = 1000/3600 m/s
u = 75.0 km/h * (1000/3600) = 20.83 m/s (rounded to 2 decimal places)
Step 2: Calculate the acceleration.
We know that the deceleration is caused by the frictional force between the tires and the road. The frictional force can be calculated using the equation:
Frictional force = coefficient of friction * normal force
The normal force is equal to the weight of the car:
Normal force = mass * gravitational acceleration (g)
Gravitational acceleration (g) is approximately 9.8 m/s^2.
Normal force = 1550 kg * 9.8 m/s^2 = 15190 N (rounded to the nearest whole number)
Now, we can calculate the frictional force:
Frictional force = 0.709 * 15190 N = 10755.31 N (rounded to 2 decimal places)
Since the frictional force is responsible for the deceleration, we can use Newton's second law:
Force = mass * acceleration
Rearranging the equation, we get:
Acceleration = Force / mass
Acceleration = 10755.31 N / 1550 kg = 6.94 m/s^2 (rounded to 2 decimal places)
Step 3: Plug the values into the equation to find the distance traveled:
s = (v^2 - u^2) / (2a)
s = (0^2 - (20.83)^2) / (2 * -6.94)
s = (-434.47) / (-13.88)
s = 31.26 m (rounded to 2 decimal places)
Therefore, the car travels approximately 31.26 m before coming to rest.
Now, let's move on to part (b) - calculating the time it takes for the car to stop.
We can use the equation:
v = u + at
Where:
v = final velocity (0 m/s when the car comes to rest)
u = initial velocity (20.83 m/s)
a = acceleration (negative, due to braking)
t = time taken
Rearranging the equation, we get:
t = (v - u) / a
Plugging in the values:
t = (0 - 20.83) / -6.94
t = 3.00 s (rounded to 2 decimal places)
Therefore, it takes approximately 3.00 seconds for the car to stop after the driver applies the brakes.