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2-6 GRAPHICAL INTEGRATION IN MOTION ANALYSIS 29

2-6GRAPHICAL INTEGRATION IN MOTION ANALYSIS


After reading this module, you should be able to...


2.18Determine a particle’s change in velocity by graphical
integration on a graph of acceleration versus time.


2.19Determine a particle’s change in position by graphical
integration on a graph of velocity versus time.

●On a graph of acceleration aversus time t, the change in
the velocity is given by


The integral amounts to finding an area on the graph:





t 1

t 0

adt


area between acceleration curve
and time axis, fromt 0 tot 1 .

v 1 v 0 


t 1

t 0

adt.

●On a graph of velocity vversus time t, the change in the
position is given by

where the integral can be taken from the graph as




t 1

t 0

vdt


area between velocity curve
and time axis, from t 0 tot 1 

x 1 x 0 


t 1

t 0

vdt,

Learning Objectives


Key Ideas


Graphical Integration in Motion Analysis


Integrating Acceleration.When we have a graph of an object’s acceleration aver-
sus time t, we can integrate on the graph to find the velocity at any given time.
Becauseais defined as adv/dt, the Fundamental Theorem of Calculus tells us that


(2-27)

The right side of the equation is a definite integral (it gives a numerical result rather
than a function),v 0 is the velocity at time t 0 , and v 1 is the velocity at later time t 1. The def-
inite integral can be evaluated from an a(t) graph, such as in Fig. 2-14a. In particular,


(2-28)

If a unit of acceleration is 1 m/s^2 and a unit of time is 1 s, then the correspon-
ding unit of area on the graph is


(1 m/s^2 )(1 s)1 m/s,

which is (properly) a unit of velocity. When the acceleration curve is above the time
axis, the area is positive; when the curve is below the time axis, the area is negative.
Integrating Velocity.Similarly, because velocity vis defined in terms of the posi-
tionxasvdx/dt, then


(2-29)

wherex 0 is the position at time t 0 andx 1 is the position at time t 1. The definite
integral on the right side of Eq. 2-29 can be evaluated from a v(t) graph, like that
shown in Fig. 2-14b. In particular,


(2-30)


If the unit of velocity is 1 m/s and the unit of time is 1 s, then the corre-
sponding unit of area on the graph is


(1 m/s)(1 s)1m,

which is (properly) a unit of position and displacement. Whether this area is posi-
tive or negative is determined as described for the a(t) curve of Fig. 2-14a.





t 1

t 0

vdt


area between velocity curve
and time axis, from t 0 tot 1 

.


x 1 x 0 


t 1

t 0

vdt,




t 1
t 0

adt


area between acceleration curve
and time axis, from t 0 tot 1 .

v 1 v 0 


t 1

t 0

adt.

Figure 2-14The area between a plotted
curve and the horizontal time axis, from
timet 0 to time t 1 , is indicated for (a) a
graph of acceleration aversustand (b) a
graph of velocity vversust.

a

t 0 t 1 t

Area

(a)

v

t 0 t 1 t

Area

(b)

This area gives the
change in velocity.

This area gives the
change in position.
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