Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 18.7 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.730. What is the speed of the automobile after 1.20 s have elapsed? Ignore the effects of air resistance.
ma=F(fr) =μmg
a= μ•g
v=v₀-at
To solve this problem, we can use the equation of motion for an object under constant acceleration:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is 18.7 m/s, and the time (t) is 1.20 s. We need to find the final velocity (v).
To determine the acceleration, we need to calculate the net force acting on the car. The only force acting in the horizontal direction is the force of friction. The equation for the force of friction is:
f = μ*N
Where:
f = force of friction
μ = coefficient of kinetic friction
N = normal force
The normal force is the force exerted by the surface to support the weight of the object. In this case, since the tires are not lifting off the ground, the normal force is equal to the weight of the car, which is given by:
N = m*g
Where:
m = mass of the car
g = acceleration due to gravity
The value of g is approximately 9.8 m/s².
Substituting the values into the equation, we have:
f = μ*N
= μ*m*g
Now, the force of friction can be related to the acceleration using Newton's second law:
f = m*a
Equating the two equations for f, we get:
μ*m*g = m*a
The mass cancels out, and we are left with:
a = μ*g
Finally, substituting the values into the equation of motion, we can solve for the final velocity:
v = u + a*t
= 18.7 + (0.730 * 9.8) * 1.20
Calculating this expression yields the final velocity (v) of the automobile after 1.20 s have elapsed.