How Does a Car's Stopping Distance Change on a Downhill Slope?

[positive infinity]

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!

 

[Hootenanny]

First of all the angle of incline would be given by $\tan^{-1}\frac{1}{10}$. A grade of 1:10 means that for every 10 meters traveled horizontally forward, the elevation drops one meter.

-Hoot

 

[positive infinity]

the angle comes out to 5.71 which is almost the same, I tried that as well and if you add 1m to every 10m the answer is always the same .1 so I'm not sure how it would make a difference if used 1:10? so I'm still confused, also does anyone know if my formula is correct?

 

[Hootenanny]

You need to start by resolving forces. While traveling down the incline a component of gravity will by acting down the slope on the car accelerating it. So you need to resolve the gravitational force so that it is parallel to the incline.

Also note that the equation for friction is $Fr = \mu R$ and not $Fr = \mu mg$, because the car is on a inclined plane, $R \not{=} mg$.

-Hoot

 

[daniel_i_l]

First find the friction coefficiant with the flat road. Then use this value to find the distance on the hill with the Vf^2 = Vi^2 + 2*a*x.