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Trigonometric functions using accel & time

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A skier races down an 18 degree ski slope. During a 5.0s interval, the

    skier accelerates at 2.5m/s squared. What are the horizontal and vertical

    components of the skiers acceleration during this time?


    2. Relevant equations

    d = ViT + 1/2aT^2
    d=distance
    Vi=initial velocity
    T=time
    a=acceleration

    3. The attempt at a solution

    I drew a right angle triangle with the hypotenuse being the 18 degree incline slope the skier is descending on. The formula above is the only relevant one I've been able to find within my notes, and that's assuming initial velocity is zero (the "skier races" bit makes me suspicious). That said, when I solve it out with the formula, I get 31.25 for the hypotenuse. From there I try to use sin to figure out the bottom side of the triangle made by the downward acceleration, but I keep getting numbers like .59 or 9, which seem far too small for that specific side in relevance to the hypotenuse. I have a feeling this is just a stupid mistake, but our teacher has yet to go over using physics equations & trigonometric functions combined. I'd be extremely grateful for a push in the right direction.
     
    Last edited: Sep 20, 2011
  2. jcsd
  3. Sep 20, 2011 #2
    Ok, good. So you knew that you needed a right triangle.

    But, you already have your hypotenuse.

    You know that the acceleration vector of the skier is 2.5 m/s^2. So, 2.5 m/s^2 is the hypotenuse, correct?

    You know that you have an 18 degree angle.

    So, what does that make a[itex]_{x}[/itex] and a[itex]_{y}[/itex]?
     
  4. Sep 20, 2011 #3
    Thanks for the quick reply! I thought that I was needing to use acceleration & time to find d then use sin/cos to find the other sides. How would I use an acceleration vector with the angle to find the other sides/accelerations? I'm also assuming that Ax and Ay are the non sloping part of the right angle triangle, right? Pardon my ignorance. This is the first physics class I've taken since 9th grade, and we still haven't gone over a whole lot in this one.
     
  5. Sep 20, 2011 #4
    Yes, Ax and Ay are the non-sloping parts of the right angle triangle.

    Do not worry about "ignorance." Ignorance would be not asking the questions. :smile:

    And I'm a student as well. So don't worry.
     
  6. Sep 20, 2011 #5
    I appreciate it. I'm unsure of how I would use an acceleration vector to find the other sides, though. Every other time I've done trigonometric functions prior to this I had an actual distance, which was why I was trying to find it as d at first.
     
  7. Sep 20, 2011 #6
    Ok, so you would simply use either cos[itex]\Theta[/itex] or sin[itex]\Theta[/itex].

    [itex]\Theta[/itex] = 18 degrees

    Now, on your right triangle, which of the non-hypotenuse sides goes up and down, and which goes left and right?

    The one that goes up and down is your a[itex]_{y}[/itex].
    The one that goes left and right is your a[itex]_{x}[/itex].

    a[itex]_{y}[/itex]=a[itex]_{t}[/itex](sin[itex]\Theta[/itex])
    a[itex]_{x}[/itex]=a[itex]_{t}[/itex](cos[itex]\Theta[/itex]) Where a[itex]_{t}[/itex] is 2.5 m/s^2.

    Correct?
     
  8. Sep 20, 2011 #7
    Alright, I believe I understand now. For a[itex]_{y}[/itex]=a[itex]_{t}[/itex](sin[itex]\Theta[/itex]), I got .77254 m/s[itex]^{2}[/itex]. I believe it'd be negative too since the acceleration is going downward for that specific part, right?

    Then for a[itex]_{x}[/itex]=a[itex]_{t}[/itex](cos[itex]\Theta[/itex]) I got 2.377 m/s[itex]^{2}[/itex]

    These feel like much better answers than what I was getting prior. I appreciate the help immensely.
     
  9. Sep 20, 2011 #8
    You are right as far as I can tell.

    Acceleration in the y-direction will be negative and will be positive in the x-direction.

    If you wish to check your answers, just do this....

    a[itex]_{t}[/itex]=[itex]\sqrt{a_{x}^{2}+a_{y}^{2}}[/itex]

    You're very welcome for the help.
     
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