A car is travelling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2. The typical reaction time for an alert driver is 0.50 s versus 2.0 s for a sleepy driver. Assuming a typical car length of 5.0 m, calculate the number of additional car lengths it takes the sleepy driver to stop compared to the alert driver. I worked this out first by using v^2=Vo^2 + 2a(Δx) to get the distance traveled 0= 33.3m/s^2 + (-9m/s^2)(x) x= 61.7m Then, I used d= vt for driver #1: (33.3m/s)(.5s)= 16.7m and for driver #2: (33.3m/s)(2.0s)= 66.7m Then, I added 61.7m to each driver's distance: driver #1 total dist traveled=78.4m driver #2 total dist traveled= 128.4m Then, subtract 78.4m from 128.4m to get the total extra distance traveled by the sleepy driver. Then, divide this figure by 5m to get total car lengths additional. 128.4m- 78.4m = 50m/5m= 10 car lengths. Now, after going through the problem it appears my initial step is really unnecessary because I could have just used d=vt for driver #1 and driver #2 and took the difference of those figures divided by 5m to get car lengths extra for the sleepy driver. Am I correct? Unless the question was asking for total distance traveled by both or either driver, then step #1 isn't even needed and this problem could have solved much quicker. Please confirm. Thanks!