How far will my car roll given a function representing the slope of the landscape I'm driving on?

So I was driving in my car thinking to myself "I wonder how far I would go (before starting to roll backwards) if I just took my foot off the brakes" I tried to figure it out myself but could not. SO I am asking the smartest people in the world, you guys!

Parameters: Write a generalized equation representing the distance a car travels based on f(x) and any other physical constants/variables needed. f(x) is a function that represents the cross section of the road you are traveling on. Thus the slope at any given point represents the angle of the road. You may treat the car as a box sliding with friction.

• What kind of friction? The dependence of frictional force on velocity can be a complicated thing. – Robert Israel Jul 3 '14 at 3:41
• @RobertIsrael Most vehicle dynamics models consider rolling resistance and drag as the two primary sources of friction. Any more complicated models are linearized, localized, and typically wrapped into the rolling resistance model. For higher-fidelity dynamics, we might consider terrain factors, etc., but for this question, that's almost certainly immaterial -- just wrap it into the rolling resistance constant. – Emily Jul 3 '14 at 4:17

This is a simple situation of force balancing. Enumerate all the forces on the car:

• Gravity: $F_g = mg\cos \theta$;
• Rolling resistance: $F_R = C_R v$;
• Drag: $F_D = C_D v^2$;
• Power: $F_A$

The sum of these forces must be zero. Drag and rolling resistance act in the opposite direction of motion. The gravity term depends on orientation. Without losing generality, we can write the equation of motion as: $$0 = F_A - F_D - F_R + F_g \\ C_R\frac{dx}{dt} + C_D \left(\frac{dx}{dt}\right)^2 - mg\cos \theta = F_A.$$

If we have no acceleration (equiv. no braking), then $F_A = 0$. Of course, $\theta$ is measured as the angle with respect to level ground. This angle may change relative to the $x$-position, so it may be more appropriate to consider the gravity term as $mg\cos \theta(x)$.

• This assumes 1-D motion, which appears to be your situation. – Emily Jul 3 '14 at 4:12
• There had better be acceleration, or I'll never get my car out of the driveway. The velocity is not expected to be constant. You need a term $m \dfrac{d^2 x}{dt^2}$. – Robert Israel Jul 3 '14 at 14:39
• That's $F_a$. If the driveway is a hill, then the acceleration is in the gravity term. – Emily Jul 3 '14 at 14:45
• @Arkamis How does one account for the fact that the car might not be in contact with the road at all (depending on its speed and the profile of the road ahead)? – user_of_math Jul 3 '14 at 16:20
• @user_of_math In such a scenario, I would bring a change of underwear. – Emily Jul 3 '14 at 16:21