I am interested in getting help with a physics problem. I need
instructions on how to solve the problem step by step.
Thanks.

========================================

An ultrasonic motion detector is used to track the motion of a ball as
it rolls down a slope.  The data      on the position versus time
graph produces a trendline: Y = 0.247X2 + 0.035X + 1.75   Compare

            your trendline to the kinematic equation xf = xi + vi t +
½ at2.  The units of displacement and time

            used by the motion detector are meters and seconds.

            a.         What is the initial position, xi, of the ball
with respect to the motion detector? _____________



            b.         What is the initial velocity of the ball, vi ? ____________



            c.         What is the acceleration of the ball, a ? ___________



            d.         Calculate the position of the cart after 2.5
seconds has passed.

---

Clarification of Question by shin181-ga on 02 Sep 2005 19:46 PDT

That's exactly how I did it, just wanted to make sure.
The only type you had was 2 * 0.247 = .594 where it actually equals .494

Thanks.

The first thing to do is to recognize the meaning of the variables in the equation:

  Y = 0.247X^2 + 0.035X + 1.75   

We are told that this equation gives the displacement as a function of
time.  That means "Y" in this equation is the position of the ball at
a given time, and "X" is the time.  Comparing this equation to the
kinematic equation, x = x_i +  v_i*t + 1/2*a*t^2 would be clearer and
easier if we use the same sympols for the variables, and reorder the
terms on the right hand side so that the powers of "t" are increasing:

 x = 1.75 +  0.035 * t + 0.247 * t^2
 x = x_i  +    v_i * t + 1/2*a *  t^2

Now equate the coefficients of terms that contain "t" raised to the
same power, while noting that the units of the experimentally
determined equation are meters and seconds:

  x_i = 1.75 meters
  v_i = 0.035 meters/second
  1/2*a = 0.247 meters/second^2, which implies that a = 2 * 0.247
meters/second^2, a = 0.594 meters/second^2

Calculating the position of the ball at 2.5 seconds is left as an
excercise to the reader.  (Substitute 2.5 seconds for "t" in the
experimentally determined equation and calculate the value of x)

I will accept that as an answer of the question btw