# Breaking on a slope

1. Apr 12, 2005

### recon

On flat ground, a driver takes 0.6 seconds to react to something signalling him to stop. When the driver hits the brakes, the car decelerates at 6 2/3 ms-2.

The driver is now driving down a long slope at 10 degrees to the horizontal. When a beam of light is flashed on him, he immediately steps on the brake (of course, he takes 0.6s to react). How long is the distance travelled before the car comes to a complete stop?

Should I resolve the acceleration due to gravity down the slope?

2. Apr 12, 2005

### Gokul43201

Staff Emeritus
What does the first part of the question tell you ?

To find the distance in the second part you must resolve all forces along the slope (including the one due to gravity).

3. Apr 13, 2005

### recon

I actually had to rephrase the question as the original question was much too long. Anyway, here is my approach to the problem:

The acceleration (due to gravity) down the slope is $$sin 10 \times 9.81ms^{-2} = 1.703488623 ms^{-2}$$.

Since he takes 0.6 seconds to react, the car has increased its speed to $$35 ms^{-1}+ (1.7 ms^{-2} \times 0.6 s) = 36.02 ms^{-1}$$ and this is the speed it is travelling at the instance before he hits the brakes.

In this time, the car has travelled a distance of $$\frac{1}{2} \times (35ms^{-1} + 36.02ms^{-1}) \times 0.6s = 21.306m$$.

We now have to calculate the distance the car traverses from the time the brakes are applied to the time when the car comes to a complete stop. The negative acceleration caused by breaking is $$6\frac{2}{3}ms^{-2}$$. Therefore the net (negative) acceleration or retardation is $$6\frac{2}{3}ms^{-2} - 1.703488623ms^{-2} = 4.963178044ms^{-2}$$.

The distance traversed from the time the brakes are applied to the time when the car comes to a complete stop is therefore =

$$\frac{1}{2} \times 36.02^2 \times \frac{1}{4.963178044} = 130.7066147m$$

So the total distance travelled from when we start observing the car to when it comes to a complete stop is $$= 130.7066147m + 21.306m \approx 152m$$.

But the answer at the back of the book says it is 146 m, and that has made me a very unhappy person.

4. Apr 13, 2005

### whozum

I can't find anything wrong with this, but when I tried using $v_0 = 35m/s$ insetad of 36.02 in the second last step, it works out to 144.7m.

Your derivation seems more correct to me though.