Question

4. How fast was a driver going if the car left skid marks that were 32 feet long on dry concrete? (The coefficient of friction is 1.02)?
73 mph


49 mph


Oo oo


38 mph


31 mph

Answer 1

Answer:

$v_{o} \approx 31.247\,mph$

Step-by-step explanation:

The deceleration experimented by the car is the product of kinetic coefficient of friction and gravitational constant:

$a = \mu_{k}\cdot g$

$a = (1.02)\cdot \left(32.174\,\frac{ft}{s^{2}} \right)$

$a = 32.817\,\frac{ft}{s^{2}}$

The initial speed is computed from the following kinematic expression:

$v^{2} = v_{o}^{2} - 2\cdot a \cdot \Delta s$

$v_{o} = \sqrt{v^{2}+2\cdot a\cdot \Delta s}$

$v_{o} = \sqrt{\left(0\,\frac{ft}{s} \right)^{2}+2\cdot \left(32.817\,\frac{ft}{s^{2}} \right)\cdot (32\,ft)}$

$v_{o}\approx 45.829\,\frac{ft}{s}$

$v_{o} \approx 31.247\,mph$