Kinematics - which equation to use in kinematics problems?

[kathyt.25]

Homework Statement

(a) A skier is gliding at 6.75m/s on horizontal, frictionless snow. He starts down an 11deg incline. His speed at the bottom is 21.7m/s. what's the length of the incline?

(b) How long does it take him to reach the bottom?

I just need help with part (b)

Homework Equations

a = delta_v / delta_t
d=0.5(vf+vi)t

The Attempt at a Solution

From part (a), I figured out that a=1.67m/s/s: a=gsin(theta)=1.87m/s/s

For part (b), how do I know what kinematics equation to use?
1 - If I use t = vf-vi /a = 7.99s, I get the correct answer.
2 - If I use d=0.5(vf+vi)t, t = 144/0.5(21.7+6.75) = 10.12s, I get the wrong answer.

How do I know which equation to use in kinematics problems? All pieces of information are available!

 

[Cyosis]

d=0.5(vf+vi)t

Is this supposed to mean the average velocity times time? Calculating averages like you do only works for linear equations. That is it would have worked if v is constant, but it is not since there is an acceleration.

 

[Doc Al]

For part (b), how do I know what kinematics equation to use?
1 - If I use t = vf-vi /a = 7.99s, I get the correct answer.

OK, you found the time. Done!

2 - If I use d=0.5(vf+vi)t, t = 144/0.5(21.7+6.75) = 10.12s, I get the wrong answer.

There's nothing wrong with using that equation, but where did you get a distance of 144? If that was your answer to part a, you'd better show how you got it.

 

[notnimdab2009]

hi, sorry for reviving this thread.

As for the computing for the length of the incline, it is correct to breakdown the g and 6.75 m/s initial velocity into components parallel to incline? Then use the formula vf^2=Vo^2 + 2aS to get the S?

 

[Doc Al]

hi, sorry for reviving this thread.

As for the computing for the length of the incline, it is correct to breakdown the g and 6.75 m/s initial velocity into components parallel to incline? Then use the formula vf^2=Vo^2 + 2aS to get the S?

That's a perfectly fine way to solve for the distance traveled down the incline. Note that the 6.75 m/s initial velocity is already parallel to the incline.

 

[notnimdab2009]

Thanks, however i find it the g parallel to the incline as g/sin11 = 51.36 /sec^2 but from the op the g is gsintheta = (9.8)(sin11) = 1.87 m/sec^2. I'm confused which is the correct g parallel to the incline that will add up to the initial speed/velocity of the skier.

 

[Doc Al]

Thanks, however i find it the g parallel to the incline as g/sin11 = 51.36 /sec^2 but from the op the g is gsintheta = (9.8)(sin11) = 1.87 m/sec^2. I'm confused which is the correct g parallel to the incline that will add up to the initial speed/velocity of the skier.

The acceleration down the incline is g sinθ. Read this: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm"

 

[notnimdab2009]

oops.. i think i got myself confused... i was getting the g prallel to the inclined when the skier is still in horizontal. So for all these types of problems, we will treat all initial velocity and the g to be parallel to the inclined already? Thanks for showing me this. I am trying to refresh my memory as i am trying to help my son in his high school physics.

 

[Doc Al]

So for all these types of problems, we will treat all initial velocity and the g to be parallel to the inclined already?

Since he's moving down the incline, his initial velocity is already parallel to the incline. But you need to figure out his acceleration, which is the component of g parallel to the incline = g sinθ.

 

[Selwyn]

one should not rely solely on formulas but on concepts
there is an accelaration,u know the length,so u can find the time taken easily
using(v^2=u^2+2XaXs)

 

[zgozvrm]

That's a perfectly fine way to solve for the distance traveled down the incline. Note that the 6.75 m/s initial velocity is already parallel to the incline.

I'm not seeing this...

If his horizontal velocity is 6.75 m/s before reaching the incline and his vertical velocity is 0 m/s (since he is initially on "horizontal, frictionless snow"), then how can his velocity parallel to the incline also be 6.75 m/s?

This is no different then having a ball rolling on a flat (horizontal) frictionless table top that drops off the edge to the ground; the initial vertical velocity at the time the ball drops is 0 m/s.

 

[zgozvrm]

... and, by my calculations, the skier first makes contact with the slope approx. 1.84 m down the hill. He makes contact at a velocity of 7.24 m/s at -21.24 degrees. This means he initially makes contact with the hill at an angle of 10.244 degrees with respect to the slope.

 

[Doc Al]

I'm not seeing this...

If his horizontal velocity is 6.75 m/s before reaching the incline and his vertical velocity is 0 m/s (since he is initially on "horizontal, frictionless snow"), then how can his velocity parallel to the incline also be 6.75 m/s?

This is no different then having a ball rolling on a flat (horizontal) frictionless table top that drops off the edge to the ground; the initial vertical velocity at the time the ball drops is 0 m/s.

I understand what you're saying and you have a good point. Nonetheless, I suspect that for simplicity they want you to assume that the skier smoothly transitions to the incline at its top. So can treat the problem as if the skier were at the top of the incline with the given speed moving parallel to the surface.

But treating the skier as a projectile breaking contact with the surface is more fun! But that makes the problem more challenging, so I don't think that was the intention. This was meant as a simpler exercise.

 

[zgozvrm]

I understand what you're saying and you have a good point. Nonetheless, I suspect that for simplicity they want you to assume that the skier smoothly transitions to the incline at its top. So can treat the problem as if the skier were at the top of the incline with the given speed moving parallel to the surface.

But treating the skier as a projectile breaking contact with the surface is more fun! But that makes the problem more challenging, so I don't think that was the intention. This was meant as a simpler exercise.

... so the question would have been better stated as such:

A skier is gliding at 6.75m/s down an 11 degree incline on frictionless snow when he passes a tree. At the bottom of the incline, his speed has increased to 21.7m/s. What is the length of the incline from the tree to the bottom of the hill?

 

[Doc Al]

... so the question would have been better stated as such:

A skier is gliding at 6.75m/s down an 11 degree incline on frictionless snow when he passes a tree. At the bottom of the incline, his speed has increased to 21.7m/s. What is the length of the incline from the tree to the bottom of the hill?

Definitely.