# How Far Will a Car Coast Up a Hill After Running Out of Gas?

• Julia Coggins
In summary, a car traveling at 30 m/s runs out of gas and coasts for 2.8s before rolling back down a 20 degree slope. The car will eventually roll all the way back down to the bottom of the slope.
Julia Coggins

## Homework Statement

A car traveling at 30 m/s runs out of gas whiles traveling up a 20 degree slope. How far up the hill will it coast before staring to roll back down?
Vf=0
Vo= 30sin20
a = -9.8sin(20)

## Homework Equations

y=yo + Vo(t) + aT^2
vf=vo+at
Vf^2=Vi^2 + 2aY

## The Attempt at a Solution

I tried finding the time the car rolls forward by using the second equation first:
0=30sin20 + (-9.8sin20)t
Solving:
t=2.80
Then, I tried solving the vertical distance using either equation 1 or 3, both unsuccesfully.
s=(30sin20)(2.8) + (9.8sin20)(2.80^2)
giving: y= 55.007 y/sin20= 160.83
or yf^2=yo^2 + 2ax
0^2=30sin20^2 + 2(-9.8sin20)(x)
giving: y= 1.5506 or d= 4.475

Obviously, both of the attempted solutions are way off base of the answer, 132 m. I don't know where I went wrong, am I using the wrong equations or did I somehow plug in the numbers wrong?

Julia Coggins said:

## Homework Statement

A car traveling at 30 m/s runs out of gas whiles traveling up a 20 degree slope. How far up the hill will it coast before staring to roll back down?
Vf=0
Vo= 30sin20
a = -9.8sin(20)

## Homework Equations

y=yo + Vo(t) + aT^2
vf=vo+at
Vf^2=Vi^2 + 2aY

## The Attempt at a Solution

I tried finding the time the car rolls forward by using the second equation first:
0=30sin20 + (-9.8sin20)t
Solving:
t=2.80
Then, I tried solving the vertical distance using either equation 1 or 3, both unsuccesfully.
s=(30sin20)(2.8) + (9.8sin20)(2.80^2)
giving: y= 55.007 y/sin20= 160.83
or yf^2=yo^2 + 2ax
0^2=30sin20^2 + 2(-9.8sin20)(x)
giving: y= 1.5506 or d= 4.475

Obviously, both of the attempted solutions are way off base of the answer, 132 m. I don't know where I went wrong, am I using the wrong equations or did I somehow plug in the numbers wrong?

Note that you have a car traveling at 30m/s and, by your calculations, it stops in 2.8s. That's a deceleration of more than 10m/s/s. You need to start sanity checking your answers, because clearly it must take much longer than 2.8s to stop.

You used ##30sin20## as your initial speed. That's going to reduce your initial speed to about 10m/s. That can't be right. The initial ##30m/s## is the speed up the slope.

To emphasise the point. If the car was moving along the flat, then the angle is ##0## and ##30sin0 =0## so the car wouldn't be moving at all!

I redid my calculations, reaching 137m instead of 134. I used 30sin20 to find the distance of y it coasts and to stay consistent with the acceleration I used, 9.81sin(20), then used trig to find the real distance.
0=30sin(20) + (-9.81)t
t=3.058
y=30sin20(3.058) + 1/2 (9.8sin20)(3.058^2)
y= 31.376 + 15.688=47.065
y/sin(20)=137 m.

Julia Coggins said:
I redid my calculations, reaching 137m instead of 134. I used 30sin20 to find the distance of y it coasts and to stay consistent with the acceleration I used, 9.81sin(20), then used trig to find the real distance.
0=30sin(20) + (-9.81)t
t=3.058
y=30sin20(3.058) + 1/2 (9.8sin20)(3.058^2)
y= 31.376 + 15.688=47.065
y/sin(20)=137 m.

I'm sorry to say those calculations make no sense to me. I think you are imagining the car is projected into the air and calculating the time to reach its highest point. Although how you get ##t = 3.058s## from your first equation is also a mystery.

Can you draw a diagram for yourself of what is actually happening to the car?

In the equation to get time, I meant 9.8sin20 not 9.8. I can do so, and it makes sense, but looking at the calculation I fail to understand why I got 137 not 134. Perhaps they used significant figures to round the numbers?

Julia Coggins said:
In the equation to get time, I meant 9.8sin20 not 9.8. I can do so, and it makes sense, but looking at the calculation I fail to understand why I got 137 not 134. Perhaps they used significant figures to round the numbers?

If a car is going at ##30m/s## and stops in ##3s##, how can it possibly go further than ##90m##? In fact, with constant deceleration it would go ##45m##.

All of your equations and calculations are misguided. You need to rethink the problem and how your equations relate to what is actually happening to the car.

Julia Coggins said:
A car traveling at 30 m/s runs out of gas whiles traveling up a 20 degree slope.
Car's velocity is 30m/s up along the incline. Why did you take it as 30sin20? Draw a diagram as PeroK has suggested. It will be really helpful.

To find the y component and then calculate the real distance traveled up the incline. I used sin(20) to find the vertical distance, then used trig to get the whole distance traveled. Managed to account for those missing three meters using significant figured, all is resolved. Thanks to everyone

Julia Coggins said:
Managed to account for those missing three meters using significant figured,
I believe 3 meters is too large to be a significant digit error (134 m is an exact answer, not rounded off).
Julia Coggins said:
To find the y component
Of what? The velocity is given up along the incline. There is no need to split it into components.
If you really think all is resolved, disregard this post.
Edit: Ok. I got your method after re-reading #5. You should get 134.25 m in the end and not 137 m.

## 1. How does the weight of the car affect how far it will coast?

The weight of the car does not have a significant impact on how far it will coast. The main factors that affect coasting distance are the initial speed and the slope of the surface the car is rolling on.

## 2. What is the average coasting distance for a car?

The average coasting distance for a car is around 50-60 meters, depending on the initial speed and the slope of the surface. However, this can vary greatly depending on the specific car and conditions.

## 3. Does the type of tires on the car affect the coasting distance?

Yes, the type of tires on the car can affect the coasting distance. Tires with more rolling resistance, such as winter tires, will cause the car to lose momentum faster and decrease the coasting distance. Tires with less rolling resistance, such as high-performance tires, will allow the car to coast further.

## 4. Can the shape or aerodynamics of the car affect how far it will coast?

Yes, the shape and aerodynamics of the car can have an impact on coasting distance. Cars with a more streamlined shape and lower drag coefficient will experience less resistance and therefore can coast further. However, the difference in coasting distance may be minimal compared to other factors.

## 5. Is it possible for a car to coast indefinitely?

No, a car cannot coast indefinitely. Eventually, the car will come to a stop due to the effects of friction and air resistance. The distance it can coast will depend on the initial speed, slope, and other factors mentioned above.

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