Vi = 100km/h Vo=0km/h incline 15°. how far?

[twenty5]

Homework Statement

I'll just sum the question up:

A vehicle is on an incline of 15° and has velocity initial of 100km/h or 27.77
m/s and is left in neutral to roll up the hill -which is inclined at 15°. How far will it roll up the hill?

Homework Equations

I tried going with V_f - V_i = d / t

but I'm not even given time in the question... T_T

The Attempt at a Solution

I absolutely need some help here. If someone can just direct which direct to go first that would be great. thanks!

 

[twenty5]

Im thinking this has to do with conservation of energy... but I don't have a mass :/

 

[Pollux]

You can figure out the time it takes it to stop, what is the velocity when it has reached the farthest up the hill?

what kind of acceleration affects the vechicle?

 

[twenty5]

You can figure out the time it takes it to stop, what is the velocity when it has reached the farthest up the hill?

what kind of acceleration affects the vechicle?

I could? let's see... V_f - V_i = d / t

I don't have the distance though... Vf = 0
Vo = 27.77m/s

gravity would affect the vehicle. So I would use A = V/t

Now, I would substitute V =d/t into A = V/t and I would end up with t... would this be correct?if so, how will the 15° come into play?

 

[twenty5]

well I diid all of that and I 2.83 seconds. I used that and stuck it into v = d/t and I got d = 78.59m. Can you please check to see if this is the correct answer? please and thank you ^^

 

[Pollux]

Yes you need to use that we have an incline

I started out with the formula: v=vi+at, do you know this formula?

 

[twenty5]

I got it to d=152m, and I think that's correct. Yes you need to use that we have an incline, I used it for the acceleration.

I started out with the formula: v=vi+at, do you know this formula?

would that be

A = (Vf - Vi) / t


AT + Vi = Vf


how would one use the incline part?

since gravity is going straight down vertically, there wouldn't really be any components other than 9.81 :S

 

[Pollux]

Acctualy you can split the gravitational acceleration up into its components, so how much of that acceleration is in the direction of vi?

Make a picture and you can see how to get its components.

 

[twenty5]

I made a drawing but it's point straight down :/

 

[twenty5]

the only other way I would get a component for the gravitational acceleration would be if I had made it perpendicular to the 15° incline :(

 

[Pollux]

Well you can decompose the accleration in a component that is perpendicular to the 15 degrees and a component that in the opposit direction of the initial velocity.

The perpendicular component would be g*cos 15, but you want the other one that is in the direction of motion.

About the formula befor. It comes from that

dv/dt=a

http://en.wikipedia.org/wiki/Kinematics#Kinematics_of_constant_acceleration here you can see how its found.

 

[twenty5]

mmk. Because I was taught that the gravitational acceleration (g) is always doing straight down :) but I'll give it a go with it being perpendictular to the 15°. thank you very much my friend ^^

 

[Pollux]

The gravitational acceleration is allways pointing down towards Earth center of mass, but that does not mean we can't split it up into components, for exampel if you want to split it up into its xyz components were z points up you would get a=(0,0,-g).

In this case we want to split it up into a component that's perpendicular and a component that's in the direction of the motion, for this we get that

a=(-g*sin 15,g*cos 15)

were the last entry is for the perpendicular component. You could write out the initial velocity in this basis as well as vi=(vi,0).

Now we just care about the motion that goes on up and down the hill, simply because we know that this is the only place were motion happens because the problem is stated as such. You then get that

v=vi-(g*sin 15)*t.

 

[Integral]

Seems to me that the easiest approach would be by setting the ke = pe , the mass will cancel so you don't need it. You have the inital velocity, solve for height, use trig to find the distance you need,