# Resolve Acceleration Due to Gravity on Slope: How Far Will the Car Travel?

• recon
In summary, 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 car travels a distance of 152m before coming to a complete stop.
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 traveled before the car comes to a complete stop?

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

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).

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 traveling at the instance before he hits the brakes.

In this time, the car has traveled 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 traveled 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.

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.

## 1. What is acceleration due to gravity on a slope?

Acceleration due to gravity on a slope is the rate at which an object speeds up or slows down as it moves down a slope due to the force of gravity.

## 2. How is acceleration due to gravity on a slope calculated?

Acceleration due to gravity on a slope can be calculated by multiplying the acceleration due to gravity (9.8 m/s²) by the sin of the angle of the slope.

## 3. How does the slope of the surface affect acceleration due to gravity?

The steeper the slope of a surface, the greater the acceleration due to gravity will be.

## 4. How can the distance traveled by a car on a slope be determined?

The distance traveled by a car on a slope can be determined using the formula: d = 1/2 * a * t², where d is the distance, a is the acceleration due to gravity on the slope, and t is the time it takes for the car to travel the distance.

## 5. Are there any other factors that can affect the acceleration due to gravity on a slope?

Yes, other factors that can affect the acceleration due to gravity on a slope include the mass and shape of the object, air resistance, and any other external forces acting on the object.

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