Mathematical Tools for Physics - Department of Physics - University

(nextflipdebug2) #1
2—Infinite Series 40

vy(t) =v 0 −gt, y(t) =v 0 t−


1

2

gt^2 (2.32)


Should you expect this? Not if you remember that there’s air resistance. If I claim that the answers are


vy(t) =−vt+


(

v 0 +vt


)

e−gt/vt, y(t) =−vtt+ (v 0 +vt)


vt


g


[

1 −e−gt/vt


]

(2.33)


then this claim has to be inspected to see if it makes sense. And I never bothered to tell you what


the expression “vt” means anyway. You have to figure that out. Fortunately that’s not difficult in this


case. What happens to these equations for very large time? The exponentials go to zero, so


vy −→ −vt+


(

v 0 +vt


)

.0 =−vt, and y −→ −vtt+ (v 0 +vt)vt


g


vtis the terminal speed. After a long enough time a falling object will reach a speed for which the force


by gravity and the force by the air will balance each other and the velocity then remains constant.
Do they satisfy the initial conditions? Yes:


vy(0) =−vt+


(

v 0 +vt


)

e^0 =v 0 , y(0) = 0 + (v 0 +vt)


vt


g


.(1−1) = 0

What do these behave like for small time? They ought to reduce to something like the expressions
in Eq. (2.32), but just as important is to determine what the deviation from that simple form is. Keep
some extra terms in the series expansion. How many extra terms? If you’re not certain, then keep one
more than you think you will need. After some experience you will usually be able to anticipate what
to do. Expand the exponential:


vy(t) =−vt+


(

v 0 +vt


)

[

1 +

−gt


vt


+

1

2

(

−gt


vt


) 2

+···

]

=v 0 −


(

1 +

v 0


vt


)

gt+


1

2

(

1 +

v 0


vt


)

g^2 t^2


vt


+···

The coefficient oftsays that the object is slowing down more rapidly than it would have without air


resistance. So far, so good. Is the factor right? Not yet clear, so keep going. Did I need to keep terms


to ordert^2? Probably not, but there wasn’t much algebra involved in doing it, so it was harmless.


Look at the other equation, fory.


y(t) =−vtt+ (v 0 +vt)


vt


g


[

1 −

[

1 −

gt


vt


+

1

2

(

gt


vt


) 2


1

6

(

gt


vt


) 3

+···

]]

=v 0 t−


1

2

(

1 +

v 0


vt


)

gt^2 −


1

6

(

1 +

v 0


vt


)

g^2 t^3


vt


+···

Now differentiate this approximate expression forywith respect to time and you get the approximate


expression forvy. That means that everything appears internally consistent, and I haven’tintroduced


any obvious error in the process of approximation.
What if the terminal speed is infinite, so there’s no air resistance. The work to answer this is


already done. Expandinge−gt/vtfor small time is the same as for largevt, so you need only look back

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