Position+Velocity+Acceleration

= Position, Velocity and Acceleration =

x'(t) = v(t) AN D v'(t) = a(t)
media type="file" key="Position_Velocity Example.mov"

In case of a free-falling object, the acceleration due to gravity is –32 ft/sec2. The significance of the negative is that the rate of change of the velocity with respect to time (acceleration), is negative because the velocity is decreasing as the time increases. Using the fact that the velocity is the indefinite integral of the acceleration, you find that


 * [[image:http://media.wiley.com/Lux/88/39588.nce024.gif align="absmiddle"]] ||
 * [[image:http://media.wiley.com/Lux/88/39588.nce024.gif align="absmiddle"]] ||

Now, at //t// = 0, the initial velocity ( //v//0) is


 * [[image:http://media.wiley.com/Lux/90/39590.nce025.gif align="absmiddle"]] ||
 * [[image:http://media.wiley.com/Lux/90/39590.nce025.gif align="absmiddle"]] ||

hence, because the constant of integration for the velocity in this situation is equal to the initial velocity, write Because the distance is the indefinite integral of the velocity, you find that


 * [[image:http://media.wiley.com/Lux/92/39592.nce026.gif align="absmiddle"]] ||
 * [[image:http://media.wiley.com/Lux/92/39592.nce026.gif align="absmiddle"]] ||

Now, at //t// = 0, the initial distance ( //s//0) is


 * [[image:http://media.wiley.com/Lux/94/39594.nce027.gif align="absmiddle"]] ||
 * [[image:http://media.wiley.com/Lux/94/39594.nce027.gif align="absmiddle"]] ||

hence, because the constant of integration for the distance in this situation is equal to the initial distance, write

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