16. A car slams on its brakes while traveling at 80 kph. If the coefficient of friction between the tires and the pavement is 0.6, for how much time does the car skid? What if the pavement is wet, reducing the coefficient to 0.3? What if the road is snow ,-covered, reducing the coefficient to 0.12?

To determine the time it takes for the car to skid to a stop, we can use the formula derived from Newton's second law and the equations of motion. The deceleration \( a \) due to friction is given by:<br /><br />\[ a = \mu g \]<br /><br />where:<br />- \( \mu \) is the coefficient of friction,<br />- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).<br /><br />The initial speed \( v_0 \) of the car is 80 kph, which needs to be converted to meters per second:<br /><br />\[ v_0 = 80 \times \frac{1000}{3600} \approx 22.22 \, \text{m/s} \]<br /><br />Using the equation of motion \( v = v_0 + at \), where the final velocity \( v \) is 0 (since the car stops), we have:<br /><br />\[ 0 = 22.22 - (\mu g)t \]<br /><br />Solving for \( t \):<br /><br />\[ t = \frac{22.22}{\mu g} \]<br /><br />Now, calculate the time for each scenario:<br /><br />1. **Dry pavement (\( \mu = 0.6 \)):**<br /><br /> \[ t = \frac{22.22}{0.6 \times 9.8} \approx 3.78 \, \text{seconds} \]<br /><br />2. **Wet pavement (\( \mu = 0.3 \)):**<br /><br /> \[ t = \frac{22.22}{0.3 \times 9.8} \approx 7.56 \, \text{seconds} \]<br /><br />3. **Snow-covered road (\( \mu = 0.12 \)):**<br /><br /> \[ t = \frac{22.22}{0.12 \times 9.8} \approx 18.89 \, \text{seconds} \]<br /><br />These are the times it takes for the car to come to a stop under each condition.