and then as
xx 0 vavgt, (2-12)in which x 0 is the position of the particle at t0 and vavgis the average velocity
betweent0 and a later time t.
For the linear velocity function in Eq. 2-11, the averagevelocity over any time
interval (say, from t0 to a later time t) is the average of the velocity at the be-
ginning of the interval (v 0 ) and the velocity at the end of the interval (v). For
the interval from t0 to the later time tthen, the average velocity is(2-13)Substituting the right side of Eq. 2-11 for vyields, after a little rearrangement,(2-14)Finally, substituting Eq. 2-14 into Eq. 2-12 yields(2-15)
As a check, note that putting t0 yields xx 0 , as it must. As a further check,
taking the derivative of Eq. 2-15 yields Eq. 2-11, again as it must. Figure 2-9a
shows a plot of Eq. 2-15; the function is quadratic and thus the plot is curved.
Three Other Equations.Equations 2-11 and 2-15 are the basic equations for
constant acceleration;they can be used to solve any constant acceleration prob-
lem in this book. However, we can derive other equations that might prove useful
in certain specific situations. First, note that as many as five quantities can possi-
bly be involved in any problem about constant acceleration — namely,xx 0 ,v,t,
a, and v 0. Usually, one of these quantities is notinvolved in the problem,either as
a given or as an unknown.We are then presented with three of the remaining
quantities and asked to find the fourth.
Equations 2-11 and 2-15 each contain four of these quantities, but not the
same four. In Eq. 2-11, the “missing ingredient” is the displacement xx 0. In Eq.
2-15, it is the velocity v.These two equations can also be combined in three ways
to yield three additional equations, each of which involves a different “missing
variable.” First, we can eliminate tto obtain(2-16)This equation is useful if we do not know tand are not required to find it. Second,
we can eliminate the acceleration abetween Eqs. 2-11 and 2-15 to produce an
equation in which adoes not appear:(2-17)Finally, we can eliminate v 0 , obtaining(2-18)Note the subtle difference between this equation and Eq. 2-15. One involves the
initial velocity v 0 ; the other involves the velocity vat time t.
Table 2-1 lists the basic constant acceleration equations (Eqs. 2-11 and 2-15) as
well as the specialized equations that we have derived. To solve a simple constant ac-
celeration problem, you can usually use an equation from this list (ifyou have the
list with you). Choose an equation for which the only unknown variable is the vari-
able requested in the problem. A simpler plan is to remember only Eqs. 2-11 and
2-15, and then solve them as simultaneous equations whenever needed.xx 0 vt^12 at^2.xx 0 ^12 (v 0 v)t.v^2 v 02 2 a(xx 0 ).xx 0 v 0 t^12 at^2.vavgv 0 ^12 at.vavg^12 (v 0 v).24 CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Table 2-1 Equations for Motion with
Constant Accelerationa
Equation Missing
Number Equation Quantity
2-11 vv 0 at xx 0
2-15 v
2-16 t
2-17 a
2-18 v 0aMake sure that the acceleration is indeed
constant before using the equations in this table.
xx 0 vt^12 at^2xx 0 ^12 (v 0 v)tv^2 v 02 2 a(xx 0 )xx 0 v 0 t^12 at^2