Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Motion Down an Incline help

  1. Mar 21, 2005 #1
    Hi Guys,

    For a physics practical investigation I will be investigating the change in height (angle) of an inclined plane and the time taken to travel down the incline.
    Now for my hypothesis and results I am a bit stuck. For the hypothesis I will be using the following formula.

    x=ut+.5at^2

    Where:
    x=distance down the incline plane (meters)
    u=Initial velocity (m/s)
    t=time taken to travel down (seconds)
    a=acceleration down the incline plane (m/s/s meters per second per second)

    Now for our situation
    x=1.58
    u=0
    t=?
    a=gSinΦ (Φ varies with different heights)

    So now I can build a relationship between the height (or the angle of inclination) and the time taken to travel.

    Now I substitute values into the formula to get:

    x=.5at^2 (because u=0)

    Then transpose:
    2x=at^2
    t^2=(2x)/(gsinΦ)

    Now we know that the values of 2x and g (g = 9.8 = acceleration due to gravity) remain constant (well we assume they do). How do I create a direct proportionality between t and Φ ?? So I get a straight line graph when I plot the two?

    Have I gone about things the wrong way?

    Edit - Here is a graph of the angle versus time
     

    Attached Files:

    Last edited: Mar 21, 2005
  2. jcsd
  3. Mar 21, 2005 #2
    You could simply plot t vs. (sin Φ)^1/2 to give you a straight line.
     
  4. Mar 21, 2005 #3
    Hi Nylex,
    is t on the vertical axis or horizontal axis?

    I ploted (sinΦ)^1/2 but that just gives me a curved graph. Then I ploted (sinΦ)^-1/2 which gave me a sort of straight graph. Have I found proportionality? I got the excel spreadsheet if anyone needs it?
     
    Last edited: Mar 21, 2005
  5. Mar 21, 2005 #4
    Given x and g to be constants your function will become
    t^2 = (2x/g) * sin(phi)
    t = (2x/g)^.5 * sin(phi)^.5

    (2x/g)^.5 is just a scalar coefficient, don't worry about it.

    Your function is reflecting the relationship t = sin(phi)^.5, there shouldnt be a linear trend because the relationship si the square root of a periodic function, so at the least it will curve. You could try logarithms in which case

    log t = (log sin(phi))/2, that should be somewhat linear.
     
  6. Mar 21, 2005 #5
    Hi whozum,

    Is that log base to e or log base to 10?
     
  7. Mar 22, 2005 #6
    Any base will work just as long as you use the same base for both sides.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook