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Kinematics - which equation to use in kinematics problems?

  1. Jun 17, 2009 #1
    1. The problem statement, all variables and given/known data
    (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. whats the length of the incline?

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

    I just need help with part (b)

    2. Relevant equations
    a = delta_v / delta_t
    d=0.5(vf+vi)t

    3. 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!
     
  2. jcsd
  3. Jun 17, 2009 #2

    Cyosis

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    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.
     
  4. Jun 17, 2009 #3

    Doc Al

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    Staff: Mentor

    OK, you found the time. Done!
    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.
     
  5. Sep 16, 2010 #4
    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?
     
  6. Sep 16, 2010 #5

    Doc Al

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    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.
     
  7. Sep 16, 2010 #6
    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.
     
  8. Sep 16, 2010 #7

    Doc Al

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    The acceleration down the incline is g sinθ. Read this: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Sep 16, 2010 #8
    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.
     
  10. Sep 18, 2010 #9

    Doc Al

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    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.
     
  11. Oct 25, 2010 #10
    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)
     
  12. Oct 25, 2010 #11
    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.
     
  13. Oct 25, 2010 #12
    ... 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.
     
  14. Oct 25, 2010 #13

    Doc Al

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    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.
     
  15. Oct 25, 2010 #14
    ... 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?
     
  16. Oct 25, 2010 #15

    Doc Al

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    Definitely. :wink:
     
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