Thomas Calculus 13th [Solutions]

698 Chapter 9 First-Order Differential Equations
(b) If v0 2 gR , the
and
2 2
v 2 2gR
2gR
,
s
v s
since v 0 if v0 2 gR . Then
2 1/2 2 2 3/2 2 3/2 3 2
s ds 2gR dt s ds 2gR dt s 2gR t C
3 1 s 2 gR t C ; t 0
2
3/2 3 2 3/2 3/2 3 2 3/2 3
3/2
s R R 2 gR (0) C C R s 2gR t R R 2g t R
2 2 2
3/2 3 1/2 3/2 3 2gR 3/2 3v 3v
2 2R 2R 2R
ds
dt
2
2gR
0 0
R R 2g t 1 R t 1 R t 1 s R 1 t
s
2/3
v0m
(0.86)(30.84)
v0 m ( k / m)
t
36. coasting distance
0.97 k 27.343. s( t) 1 e
k
k
k
(27.343/30.84) t
0.8866t
s( t) 0.97 1 e s( t) 0.97 1 e . A graph of the model is shown superimposed
on a graph of the data.
CHAPTER 9
ADDITIONAL AND ADVANCED EXERCISES
dy dy dy
1. (a) k A c y dy k A y c dt k A dt k A dt ln y c k A t C
dt V V y c V y c V V 1
k
At
1 V
.
C
y c e e Apply the initial condition, y (0) y0 y0 c C C y0
c
k t
y c y
V
0 c e .
A
k t
(b) Steady state solution: y lim y( t) lim c y
V
0 c e c y0
c (0) c
t t
A
d mv
d mv
2. F v u dm F v u dm F m dv v dm v dm u dm F m dv u dm .
dt dt dt dt dt dt dt dt dt dt
dm b m b t C . At t 0, m m
dt
0,
so C m 0 and m m0 b t.
Thus
0 dv
u b m b t
0 dv
1
F m b t 0
ln
dt u b m b t g dt g m0
b t
v gt u m
C
0
m0 b t dy
m0
b t
v 0 at t 0 C 1 0. So v gt u ln
y gt u ln dt and u c, y 0 at
m dt m
1 2 m0 b t m0
b t
t 0 y gt c t ln
2
b m
0
0 0
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