about the functionx(t), given information about the acceleration.
To go from acceleration to position, we need to integrate twice:x=∫ ∫
adtdt=
∫
(at+vo)dt [vois a constant of integration.]=∫
atdt [vois zero because she’s dropping from rest.]=1
2
at^2 +xo [xois a constant of integration.]=1
2
at^2 [xocan be zero if we define it that way.]Note some of the good problem-solving habits demonstrated here.
We solve the problem symbolically, and only plug in numbers at
the very end, once all the algebra and calculus are done. One
should also make a habit, after finding a symbolic result, of check-
ing whether the dependence on the variables make sense. A
greater value oftin this expression would lead to a greater value
forx; that makes sense, because if you want more time in the
air, you’re going to have to jump from higher up. A greater ac-
celeration also leads to a greater height; this also makes sense,
because the stronger gravity is, the more height you’ll need in or-
der to stay in the air for a given amount of time. Now we plug in
numbers.x=1
2
(
9.8 m/s^2)
(1.0 s)^2
= 4.9 m
Note that when we put in the numbers, we check that the units
work out correctly,(
m/s^2)
(s)^2 = m. We should also check that
the result makes sense: 4.9 meters is pretty high, but not unrea-
sonable.
The notation dqin calculus represents an infinitesimally small
change in the variableq. The corresponding notation for a finite
change in a variable is ∆q. For example, ifqrepresents the value
of a certain stock on the stock market, and the value falls from
qo = 5 dollars initially toqf = 3 dollars finally, then ∆q = − 2
dollars. When we study linear functions, whose slopes are constant,
the derivative is synonymous with the slope of the line, and dy/dx
is the same thing as ∆y/∆x, the rise over the run.
Under conditions of constant acceleration, we can relate velocity
and time,
a=
∆v
∆t,
or, as in the example 1, position and time,x=1
2
at^2 +vot+xo.22 Chapter 0 Introduction and Review