1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A skier starts from rest, accelerates down 30degree slope

  1. Oct 14, 2011 #1

    1. A skier starts from rest at the top of a hill of height h down a 30.0 degree ski slope with an acceleration a.

    a.) What is her speed at the bottom of the hill?
    b.) If the hill was h = 250. m high and her acceleration down the slope was a = 4.00 m/s2, what was instantaneous speed at the bottom of the hill, and her average speed down the slope?
    c.) At the base of the hill, she continues horizontally for another 250 m before coming to a stop and ending her run. What is her total vector displacement from start to finish?




    I'm asking about a) for now.

    for a.) I drew the FBD, rotated it so Fgx is parallel to the slope of the hill.
    mgSin30=Fgx
    Fgx=4.9m

    then I did Fnet/m=a
    the m's cancel so I'm left with a=4.9 m/s^2
    it's asking for the speed though, aka velocity so is my answer expected to simply be 4.9m/s^2*t? I feel like I should be able to work with it more but i don't see how I could?
     
  2. jcsd
  3. Oct 14, 2011 #2
    For part b) I used the formula Vf^2-Vi^2=2a(\delta)x

    For the change in x I found the hypotenuse of the triangle, is that correct? or should I have used the 250m that was given?

    After plugging in the values I got 63.25 m/s for my instantaneous velocity.
     
  4. Oct 14, 2011 #3
    I'm not sure how to find the average velocity. Unless all I have to do is V=at? In which case I get 31.64 m/s
     
  5. Oct 14, 2011 #4
    I know its early but can anyone else verify if the work I'm doing is correct? :(
     
  6. Oct 14, 2011 #5
    For C) For letter C is it simply 750m?
     
  7. Oct 14, 2011 #6
    Acceleration isn't velocity, you have to work out velocity. I think you need to use the suvats.
    s=displacement(m)
    u=start velocity(ms-1)
    v=end velocity(ms-1)
    a=acceleration(ms-2)
    t=time(s)
    There are 5 suvat equations, each omits one of the 5 measurements. Depending on the information given, you use a different suvat equation.
    v=u+at (no s)
    s=vt-0.5at2 (no u)
    s=ut+0.5at2 (no v)
    s=0.5t(u+v) (no a)
    v2=u2+2as (no t)
    Remember:These equations can only be used if acceleration is constant.
    Where it stops (e.g. the point turns around mid-air, or via the force of friction) then v=0.
    If it starts from rest, u=0.
    s is displacement, not velocity.
    Any one of the measurements can be negative or come out negative.
    If s or t comes as two answers, this is because it passes the same point twice.

    I gave you more than the information you need to solve this problem, but suvats will come in useful in all sorts of places, so learning how to use them is a good idea.
     
  8. Oct 14, 2011 #7
    I understand that, but velocity is equal to acceleration multiplied by time, but how do I find time? I feel like im not given enough to solve the problem, i ended up with (t)(4.9m/s^2) as my velocity

    I don't have the displacement or distance, neither do i have the time. so how can i find the velocity with such little information, unless the question itself doesn't expect me to do anything more than simplify?
     
  9. Oct 14, 2011 #8
    Yes you do.
    Even if you didn't, you can still give your answer in terms of height or time.
     
  10. Oct 14, 2011 #9
    I thought a) and b) and c) where all different questions pertaining to the same problem and unrelated, fml you might be onto something.
     
  11. Oct 14, 2011 #10
    Sometimes they make mistakes with questions and their orders. Is this an offical exam paper or just a question given by your book or teacher?
     
  12. Oct 14, 2011 #11
    A question given by my professor, I have an exam in two hours and i procrastinated looking at his review sheet. i emailed him but i doubt he'll get back to me before the time of the test. (i've been studying our past quizzes/hw instead for the past week)
     
  13. Oct 14, 2011 #12
    I still think each question is separate though i wish the question was more specific.
     
  14. Oct 14, 2011 #13
    Question: those suvat equations, are those used to find the average or instantaneous?

    What would I use to find inst. velocity vs avg velocity?
     
  15. Oct 16, 2011 #14
    Instantaneous. Average velocity is the midpoint of u and v.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A skier starts from rest, accelerates down 30degree slope
  1. Skier Going Down Slope (Replies: 12)

  2. Skier down a slope (Replies: 4)

  3. Skier going down slope (Replies: 4)

Loading...