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A very tricky problem that im just not getting

  1. Feb 13, 2006 #1
    hey all

    i have this very annoying physics problem that ive been looking at for hours.

    In the physics laboratory, a glider is released from rest on a frictionless air track inclined at an angle. If the glider has gained a speed of 22.0 cm/s in traveling 40.0 cm from the starting point, what was the angle of inclination of the track? Assume that the initial velocity is zero.

    I understand that you have the initial velocity and the final velocity, and you can find the acceleration with a "constant acceleration" equation, specifically DeltaX= 1/2acceleration*time squared, but how on earth do you find the angle of inclination given only the value for x, and not for y or the hypotenuse?

    Someone please help!!!!!!
  2. jcsd
  3. Feb 13, 2006 #2
    oh....my work

    So through using said acceleration equation
    (v squared = v initial squared plus 2(acceleration)(displacement)

    so i got that the acceleration is 6.05 cm/s

    then from that i got that the time is equal to 22= 6.05t

    So t is 3.64 seconds.

    So i'm guessing theres a way to relate the sin of the angle to the acceleration or the time?
  4. Feb 13, 2006 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    If the angle of inclination were 90 degrees, the glider would be falling straight down, accelerating at g=9.8 m/s^2. If the angle of inclination were 0 degrees, the glider would be sitting still, with an acceleration of 0.

    The acceleration is therefore equal to g times the sine of the angle of inclination.

    - Warren
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