Hey hows everyone well I'm busting my brain over this problem, On a level road the stopping distance for a certain car going 80km/hr is 32.0m. What would be the stopping distance for this car when going downhill on a 1:10 grade? A grade of 1:10 means the elevation drops 1.0m for a foward travel of 10.0m along the roadway. Take the coefficient of friction on both the level road and the hill to be the same. Now here's what I did, I found out the angle of the incline by using Sin-1 of 1\10 which is .1 which came out to be 5.74degrees. I then convereted the 80km/hr to m/s by 80km/hr*1000m\1km*1hr\60min*1min\60s = 22.2m/s. Then I used the formula vxf^2=Vxi^2+2ax(xf-xi), the final velocity and initial distance are both zero so I get 0=Vxi^2+2ax(xf). I think this is where I went wrong I added Sin(5.74) so it looks like this Vix^2+Sin(5.74)\2(μkg). The book says to convert 2ax to 2(μkg). g being gravity (9.8m/s^2). I found the coefficient of friction which came out to 0.787, So my final work was 22.2^2+Sin(5.74)\2(0.787*9.8)=32.0m which is the same in a straight as in the problem but the answer he gave us is 36.8???? Some one please help me It would be much appriciated!!!