CK-12-Physics - Intermediate

(Marvins-Underground-K-12) #1

2.5. The Kinematic Equations http://www.ck12.org


2.5 The Kinematic Equations


Objectives


The student will:



  • interpret area in an acceleration-time graph.

  • represent motion using a velocity-time graph.

  • interpret slope and area in a velocity-time graph.


Vocabulary



  • linear:Related to, or resembling a straight line.

  • midpoint:The point of a line segment or curvilinear arc that divides it into two parts of the same length; a
    position midway between two extremes.


Equations


1.~vavg=∆∆xt, always true
2.~aavg=∆∆~vt, always true
3.~vf=~at+~vi, constant acceleration only
4.~vavg=(~vf+ 2 ~vi), constant acceleration only
5.~x=^12 ~at^2 +~vit+~xi, constant acceleration only
6.~vf^2 =~vi^2 + 2 ~a∆~x, constant acceleration only

Introduction


The units associated with real world variables are a great aid in understanding physical phenomena. In this section it
will become clear why labeling variables associated with the slope, or the instantaneous slope (given by the tangent
to a curve), and area under a curve, are so useful.


Graphs of Acceleration vs Time


We begin with a car stopped at a red light. When the light turns green, the car begins to move with a constant
(uniform) acceleration of 2.0 m/s^2. We define the instant that the car begins to move ast= 0 .0 s, and the initial
position of the car, at the red light, as 0.0 m.


How is the graph of acceleration versus time over the time interval [0.0, 4.0] seconds plotted?


Since the acceleration is uniform, it remains 2.0 m/s^2 , the entire four seconds, which suggests a horizontal line for
the acceleration-time graph.


Consider the units of the area in the acceleration-time plane. Any area will do. Let’s choose a rectangular area.
We compute the rectangular area the same way we compute any rectangular area: multiply the rectangle’s length
by its width. Our interest is not in computing a number. Instead, we wish to determine the units of the area. At
first glance, this may seem a confusing goal- after all, area is measured in units of length squared, such as meters

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