Kinematics: 1-D Motion Problem with Inclined Surface - Solving for Distance

[tascja]

Homework Statement

A skier is gliding along at 3.37m/s on horizontal, frictionless snow. He suddenly starts down a 22.7(degree) incline. His speed at the bottom is 19.2m/s. What is the length of the incline?

Homework Equations

Vf^2 = Vi^2 + 2ad

The Attempt at a Solution

d = ?
Vi = 3.37 m/s
Vf = 19.2m/s
a = -9.8m/s^s


(19.2)^2 = (3.37)^2 + 2(-9.8)d
d = 18.2 m

I was wondering if this is the correct way to approach a question like this, and also why do they give the angle of the incline?

 

[Stovebolt]

Think about the direction of acceleration due to gravity. Compare this to the direction the skier is moving - they are not the same. You will need to take the component of acceleration in his direction.

 

[tascja]

I think understand what you mean, like the skier is moving down on an angle of 22.7(deg) but gravity acts straight down... only thing is that we havnt learned that yet in this class, ill give it a shot but I am wondering if there is possibly another way to do it??

 

[Stovebolt]

You're on the right track.

There really isn't any other way to do this than to divide the acceleration due to gravity into components - one component in the direction the skier is traveling, the other perpendicular to this. Since there is no friction, you will not need to use the perpendicular component of the force.

Make sure you draw out a diagram. It will help you to avoid confusion.

The best way to do this is to use trigonometric functions (sine, cosine, or tangent, depending on what is appropriate for the component you are looking for).

 

[tascja]

so when i do this, i still don't end up with the right answer, so ill show all my steps and someone can correct me if I am wrong:

3.37m/s
_______
|\ * angle: (90-22.7) = 67.3 deg
| \
|*\
| \ x
| \
| \
| \
v \______ 19.2 m/s
9.8m/s^2

so using: cos(*) = 9.8 m/s^2 / x
x = 25.3978 m/s^2

Then: Vf^2 = Vi^2 + 2ad
(19.2)^2 = (3.37)^2 + 2(25.3978)(d)
d = 7.03m
Therefore the length of the incline is 7.03m

 

[tascja]

sorry the attempt at a drawing didnt come out so nice .. lol

 

[LowlyPion]

I think it's easier to consider that the component of acceleration down the plane is given by g*sin22.7°.

That means that for the a in the expression you want to use that it would be 9.8*(.386)

 

[talaroue]

I was in the same boat as you I wasn't shown how to work the acceleration components but I had to use it. What I did was I found the acceleration components for the x and y direction. The used this equation... delta Y= final velocity^2 - inital velocity^2/(2*acceleration component for y) then i did the same for the x direction just using the x component. Then I justed used delta x^2 +delta y^2= hypotnese^2. Then i got the right awnser.

How would I find how long it takes him to reach the bottom?

 

[LowlyPion]

I was in the same boat as you I wasn't shown how to work the acceleration components but I had to use it. What I did was I found the acceleration components for the x and y direction. The used this equation... delta Y= final velocity^2 - inital velocity^2/(2*acceleration component for y) then i did the same for the x direction just using the x component. Then I justed used delta x^2 +delta y^2= hypotnese^2. Then i got the right awnser.

How would I find how long it takes him to reach the bottom?

Welcome to PF.

You should know what the acceleration is, and you should know the distance, so how would you relate distance, acceleration and time?

 

[Lopina]

Well, I'd solve it this way

\frac{mv_{1}^{2}}{2}+mgh=\frac{mv_{2}^{2}}{2}

\frac{v_{1}^{2}}{2}+gh=\frac{v_{2}^{2}}{2}

h=\frac{v_{2}^{2}-v_{1}^{2}}{2g}

d=\frac{h}{sin(\alpha)}

 

[talaroue]

What acceleration do I use though? Since its on an angle moving in the x direction would I use the 9.8cos(angle) as my acceleration?

 

[talaroue]

nvm I use the acceleration that is parellel with the slope so in this case it is the acceleration y component! If anyone is having problems I will send you my work via pm or email.