a driver sees a horse on the road and applies the brakes so hard that they lock and the car skids to a stop in 24 m. the road is level and the coefficient of kinetic friction between the tires and the road is 0.7. How fast was the car going when the brakes were applied?
Vf^2=Vi^2+2ad
but a=force/mass=mu*mg/m=mu*g
solve for vi
To determine the speed of the car when the brakes were applied, we can use the principles of kinematics and the relationship between the stopping distance, coefficient of kinetic friction, and initial velocity.
First, let's define some variables:
- Stopping distance (d) = 24 m
- Coefficient of kinetic friction (μ) = 0.7
- Initial velocity (v) = unknown
Now, we can use the equation of motion to relate the variables. The equation for the stopping distance is given by:
d = (v^2) / (2μg)
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Rearranging this equation, we can solve for the initial velocity:
v^2 = 2μgd
v = √(2μgd)
Plugging in the known values:
v = √(2 * 0.7 * 9.8 * 24)
v ≈ √(329.76)
v ≈ 18.16 m/s
Therefore, the car was going approximately 18.16 m/s when the brakes were applied.