# How far will the car coast?

1. Apr 28, 2016

### Julia Coggins

1. The problem statement, all variables and given/known data
A car travelling at 30 m/s runs out of gas whiles travelling 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)

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

3. 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?

2. Apr 28, 2016

### PeroK

Note that you have a car travelling 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!

3. Apr 28, 2016

### Julia Coggins

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.

4. Apr 28, 2016

### PeroK

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?

5. Apr 28, 2016

### Julia Coggins

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?

6. Apr 28, 2016

### PeroK

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.

7. Apr 28, 2016

### cnh1995

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.

8. Apr 28, 2016

### Julia Coggins

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

9. Apr 28, 2016

### cnh1995

I believe 3 meters is too large to be a significant digit error (134 m is an exact answer, not rounded off).
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.