Question

In the braking test of a sports car, its velocity is reduced from 70 mi/h to zero in a distance of 180 ft with slipping impending. The coefficient of kinetic friction is 80 percent of the coefficient of static friction.(b) the stopping distance forteh same initial velocity if the car skids. Ignore airresistance and rolling resistance.

Answer 1

Answer:

The stopping distance if the car skids is 225 ft

Explanation:

Friction

It's a force that opposes movement and requires the interaction of two surfaces. If the interaction occurs from relative rest, then the friction force is greater than the case where the interactions occur from relative speed. The static coefficient is greater than the kinetic friction, and the relation is

$\mu_k=0.8\mu_s$

If the car needs 180 ft to stop with slipping impending (no relative speed between the tires and the road), we can find the distance needed to stop the car in skidding conditions, which we expect to be greater.

This indicates that the friction forces have the same relation

$F_{rk}=0.8F_{rs}$

Since the mass is the same:

$m.a_{k}=0.8\cdot m.a_{s}$

Simplifying

$a_{k}=0.8\cdot a_{s}$

Now we'll focus on the dynamic formulas. The acceleration can be computed from the initial speed vo, the final speed vf and the distance x:

$\displaystyle a=\frac{v_f^2-v_o^2}{2x}$

This relation stands for both accelerations, which happen to be decelerations:

$\displaystyle a_k=\frac{v_f^2-v_o^2}{2x_k}$

$\displaystyle a_s=\frac{v_f^2-v_o^2}{2x_s}$

Where xk and xs are the distances needed to stop the car in each case. Note that vf and vo are the same since the test is done with the same values for both. Knowing the relation between the accelerations, we can have the relationship between the distances

$\displaystyle \frac{v_f^2-v_o^2}{2x_k}=0.8\cdot \frac{v_f^2-v_o^2}{2x_s}$

Simplifying

$\displaystyle x_k=\frac{x_s}{0.8}$

Thus

$\displaystyle x_k=\frac{180}{0.8}$

$x_k=225\ ft$