Question

A driver slammed on her brakes and came to a stop with constant acceleration. Measurements on her tires and skid marks on the pavement indicate that she locked her car's wheels, the car traveled 58.52 meters before stopping, and the coefficient of kinetic friction between the road and her tires was 0.750. How fast was she driving

Answer 1

Answer:

u = 29.22 m/s

Explanation:

distance (s) = 58.52 m

coefficient of kinetic friction (k) = 0.75

final velocity (v) = 0 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

How fast was she driving (u)

we can get how fast she was driving by using the formula below

s = ut - $\frac{1}{2}.at^{2}$  ...equation 1

where

• s = distance
• u = her initial velocity
• a = acceleration = $\frac{f}{m} = \frac{kmg}{m} = kg$

• k = coefficient of kinetic friction
• g = acceleration due to gravity

• t = time

        from v = u - at  (recall that v = 0)

        0 = u - at, therefore t = u/a = u/kg

now substituting the required values above into equation 1 we have

s = $\frac{u^{2}}{kg} - \frac{u^{2}}{2kg}$

s = $\frac{u^{2}}{2kg}$

u = $\sqrt{2kgs}$

u = $\sqrt{2 x 0.75 x 9.8 x 58.52}$

u = 29.22 m/s