# Homework Help: Coefficients for the equation of the position vs time equation

1. Aug 25, 2012

### Neutrinogun

EDIT: I figured it out by looking at this link pages 65-66. Thanks for looking though! http://www.bfasta.net/assets/files/...46 Information/HSU/Chapter 4 Acceleration.pdf

1. The problem statement, all variables and given/known data
Recently I just did a physics lab for kinematics in which we found the position, speed, and acceleration as time passed of a moving object. I finished most of the lab questions, however am curious about 2 aspects of the equation.

2. Relevant equations

Position vs time graph equation: $$y = 0.5424x^2 + 0.2072x + 0.0149$$

3. The attempt at a solution
One of the questions asks the relation of the coefficient of the x^2 term to the graph of the speed vs time and the graph of the acceleration vs time. I figured out that it's 1/2 the slope of the speed vs time/equivalent to the acceleration, so I finished the question. However, I have no clue why this is true (this isn't part of the lab question, I'm just curious).

Another question asks how the coefficient of the x-term relates to the graph of the speed vs time and acceleration vs time. I already figured out that it is equivalent to the y-intercept of the speed vs time graph, but am not sure if it relates to the acceleration vs time graph. The only relation I can see is that it is about 1/5 of the y-intercept of the acceleration equation. Is this correct? (And why, if it is).

EDIT: I figured it out by looking at this link pages 65-66. Thanks for looking though! http://www.bfasta.net/assets/files/...46 Information/HSU/Chapter 4 Acceleration.pdf

Last edited: Aug 25, 2012
2. Aug 25, 2012

### CWatters

I assume "x" in that equation is "time"? In which case..

If that's the equation for position then differentiate it and you get the equation for the velocity. Differentiate it again and you get the equation for acceleration.

I suggest doing that and then plotting the graphs. The coefficients may make more sense.

For example consider the standard straight line equation

y=mt + c

m is the slope of the "curve". Differentiate wrt t and you get

dy/dt = m

but dy/dt is a velocity so the coefficient m is the velocity.

Last edited: Aug 25, 2012