**Scale of the Line Rider**

First, we assume that the line rider is on Earth and for low speeds will have a free-falling acceleration of 9.8 m/s^{2}. Next, an arbitrary distance is selected. In this case the length of the sled is chosen to be 1 LU (Linerider Unit).

![line rider](http://blog.dotphys.net/wp-content/uploads/2008/09/line-rider.jpg)

The goal will be to put the linerider in a free fall (where air resistance should be able to be ignored) and determine his (it could be a she, it is difficult to tell) acceleration in LU/s^{2}. Then we can determine the conversion factor from LU/s^{2} to m/s^{2}.

For this measurement, we create a track that launches the rider nearly vertical. At the top of his trajectory, he will have a low speed (so little air resistance) and his acceleration can be measured.

![linerider](http://blog.dotphys.net/wp-content/uploads/2008/09/linerider.jpg)

Notice in this track, there are extra lines. These are used to track how the background moves while the game is running.

This trajectory produces the following y-position vs. time graph.

![linerider](http://blog.dotphys.net/wp-content/uploads/2008/09/linerider1.jpg)

This graph show the y-position vs. time for a point on the linerider. The point used was on his front right at the transition from his white shirt to his black pants (again, it could be a girl there is just not enough evidence to say one way or the other). Ideally, the center of mass should be used (this could be close).

Why should the trajectory be a parabola? Let’s start with two things, the definition of velocity and the definition of acceleration. (In this case, we will discuss these as scalar components in the y-direction:

![linerider](http://blog.dotphys.net/wp-content/uploads/2008/09/linerider2.jpg)

Where really, this is the avg velocity during the time interval delta t. If we want to get an expression for y:

![linerider](http://blog.dotphys.net/wp-content/uploads/2008/09/linerider3.jpg)

If an object is in free fall, the only force acting on it is the gravitational force. This will give an acceleration in the y-direction of -9.8 m/s^{2}. The definition of acceleration (in the y-direction) is:

![linerider](http://blog.dotphys.net/wp-content/uploads/2008/09/linerider4.jpg)

Solving for the final y velocity:

![Linerider](http://blog.dotphys.net/wp-content/uploads/2008/09/linerider5.jpg)

Now all of this can be put back into the expression for the final y position, where

![Linerider 2](http://blog.dotphys.net/wp-content/uploads/2008/09/linerider-2.jpg)

Putting this in for v_{avg}:

![yf](http://blog.dotphys.net/wp-content/uploads/2008/09/yf.jpg)

And now substituting for the final velocity:

![yf2](http://blog.dotphys.net/wp-content/uploads/2008/09/yf2.jpg)

This gives the quadratic relationship between y and time (the other values do not change)

![yf3](http://blog.dotphys.net/wp-content/uploads/2008/09/yf3.jpg)

And POOF!!!! That is the kinematic equation you remember from high school. Of course, your teacher probably told you that you needed calculus to derive it, but you don’t. The only assumption made was that the velocity changed at a constant rate (the acceleration was constant). This allowed us to say that the average velocity was the final velocity plus the initial velocity divide by 2. Ok, I lied. You probably didn’t see the equation in the form above. You probably saw something like this:

![yt](http://blog.dotphys.net/wp-content/uploads/2008/09/yt.jpg)

This says y is a function of time (which it is) and instead of delta t, this equation just has t (which is true if you let t_{0} = 0 seconds. One other change is the -g for the acceleration.

The point I am trying to make is that for constant acceleration, the position as a function of time should be a quadratic relationship.

**Back to the Scale**

So, from this data, the linerider has a y-acceleration of

![accel](http://blog.dotphys.net/wp-content/uploads/2008/09/accel1.jpg)

Where A is the coefficient of the t^{2} term, thus it must be equal to 1/2 a.

so

![LU](http://blog.dotphys.net/wp-content/uploads/2008/09/lu.jpg)

and

![1LU](http://blog.dotphys.net/wp-content/uploads/2008/09/1lu.jpg)

**Then how big is the line rider?**

After making the rider crash, he can be seen stretched out. From this, his length can be determined:

![How big is the linerider](http://blog.dotphys.net/wp-content/uploads/2008/09/how-big-is-the-linerider.jpg)

In this case the measurement is in meters. 1.116 meters is about 3.7 feet. This person is probably not a grown individual.

[According to this growth chart](http://www.keepkidshealthy.com/growthcharts/boystwoyears.gif), a 5 year old boy is about 1.1 meters tall. [A girl this height](http://www.keepkidshealthy.com/growthcharts/girlstwoyears.gif) would be about 5.5 years old. Needless to say, the line rider is either extremely short, or about 5 years old. I have a 5 year old child and I would not let him ride this sled on these user-created lines. It is just too dangerous.

**Summary**

The length of the linerider’s sled is about 1 meter.

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