# Calculating stopping distance of a car on a slope

I'm having trouble setting up a homework problem properly.

We are given the following:

A given car can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What is its stopping distance on a roadway sloping downward at an angle of 26.0°, in feet?

Because the trigonometry works out even when theta is zero, I drew a freebody diagram for the general case: In the top half of the photo, we have an inclined solid line indicating the surface on which the car drives, and other marks to indicate where measurement of the car's travel distance, initial velocity and final velocity begins and ends.

In the lower half of the photo, I have a freebody diagram with the basis for the coordinate system rotated theta degrees so that the y-direction is parallel to the normal force, and the x-direction is parallel to the surface on which the car drives.

Next, I use Newton's Second Law in the y-direction to find the normal:

$$\lvert \overrightarrow N \rvert - mg\ cos \theta = 0$$ $$\lvert \overrightarrow N \rvert = mg\ \ cos \theta$$

Then, I use Newton's Second Law in the x-direction to find acceleration:

$$-\ F_b\ -\ F_k\ +\ mg\ sin \theta\ =\ ma$$ $$-\ F_b\ -\ \mu_k\ mg\ \ cos \theta\ +\ mg\ sin \theta\ =\ ma$$ $$-\ F_b\ -\ mg\ (\ \mu_k\ \ cos \theta\ +\ sin \theta\ )\ =\ ma$$ $$-\ \frac{F_b}{m}\ -\ g\ (\ \mu_k\ \ cos \theta\ +\ sin \theta\ )\ =\ a$$

Next, I use a kinematic equation to find another general formula for acceleration:

$$V_f^2\ =\ V_0^2\ +\ 2as$$ $$\frac {-V_0^2}{2s}\ =\ a$$

Then, I use another kinematic equation to provide a value for time:

$$V_f\ =\ V_0\ +\ at$$ $$\frac {V_f\ -\ V_0}{a}\ =\ t$$ $$V_f\ -\ V_0\ \frac {2s}{-V_0^2}\ =\ t$$ $$V_f\ +\ \frac {2s}{V_0}\ =\ t$$

So, now I can calculate acceleration and time for the first half of the original question--the scenario with no incline and a given distance.

But this pesky term:

$$F_b$$

the force exerted by the brakes of the car--I don't see any way to calculate that, or cancel it.

The same goes for the mass of the car: it isn't provided, I can't measure it or remove it from my equations.

So, I'm not able to find the coefficient of kinetic friction for the car, nor its mass, nor the force exerted by the brakes.

What am I missing?

• You can't solve for $F_b$ and $m$ independently, but you can solve for $\frac{F_b}{m}$, which is all that should factor into the next part.
– Chris
Feb 11, 2016 at 1:43
• Oh and tell your tutor to start using SI units! ;-)
– Gert
Feb 11, 2016 at 1:51
• Check your newton's second law in the x and y direction. In the y-direction how are you getting that extra mg? it should just be N-mg$\cos\theta$. And in the x direction you're missing mgsin$\theta$. Just use 1 force Fb instead of both Fb and Fk. Assume the same Fb acts in the horizontal case and the tilted case. Gert has given how to find Fb. Feb 11, 2016 at 2:00
• You have a kinematic equation for acceleration. Use that to find the acceleration in the horizontal case. Then Fb is just m times that acceleration. Substitute that into the angled case. Feb 11, 2016 at 2:02
• No friction coefficient is needed to solve this problem. $F_b-mg\sin\theta=ma$ is all you need.
– Gert
Feb 11, 2016 at 2:07

Solve for $F_b$ from the horizontal braking distance. Assume $F_b$ is constant, then during braking kinetic energy has been converted to friction work:
$$F_b \Delta x = \frac12 mv^2$$
where $\Delta x=123\:\mathrm{ft}$ is the braking distance and $v=60.0\:\mathrm{miles/hour}$.