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14. t 1 s; away
15. a. s1t2 kt 2 16k 2 10k2t 2k
v1t2 2kt 16k 2 10k2
a1t2 2k 0
2k
Since k 0 and kR, then
a1t2 2k 0 and an element of
the real numbers. Therefore, the
acceleration is constant.
b. t 5 3k, 9k 3 30k 2 23k.
16. a. The acceleration is continuous at
t 0 if lim a1t2 a102.
tS0
For t 0,
and
and
s1t2
t3
t 2 1
v1t2 3t2 1t 2 12 2t1t 3 2
1t 2 12 2
tS0
t4 3t 2
1t 2 12 2
a1t2 14t3 6t21t 2 12 2
1t 2 12 2
21t2 1212t21t 4 3t 2 2
1t 2 12 2
14t3 6t21t 2 12
1t 2 12 3
4t1t4 3t 2 2
1t 2 12 3
4t5 6t 3 4t 3
1t 2 12 3
6t 4t5 12t 3
1t 2 12 3
2t 3 6t
1t 2 12 3
Therefore,
0, if t 6 0
a1t2 • 2t 3 6t
1t 2 12 3 , if t 0
and
0, if t 6 0
v1t2 • t 4 3t 2
1t 2 12 2,
if t 0
lim lim
tS0 a1t2 0 tS0a1t2 0,
1
0
Thus, lim a1t2 0.
Also, a102 0 1
0
Therefore, lim a1t2 a102.
tS0
Thus, the acceleration is
continuous at t 0.
17.
18.
b. velocity approaches 1,
acceleration approaches 0
v 1b 2 2gs2 1 2
dv
dt 1 2 1b2 2gs2 1 ds
2 a 0 2g
dt b
a g
Since g is a constant, a is a constant,
as required.
ds
Note:
dt v
dv
dt a
d
F m 0
dt a
v
1 1 v c 2 2 b
Using the quotient rule,
1
2v dv
2Q1 v2
c2R 1 2 a dt b v
c2
1 v2
c 2
dv
Since
dt a,
v b 2 2gs
a 1 2v 2gv
1
dv v2
m 0 Q1
dt c 2R 2
1 v2
c 2
m 0 Q1 v2
c 2 R 1 2
c aQ1 v2
c 2 R v2 a
c 2 d
m 0 c ac2 av 2
c 2
Q1 v2
c 2 R
1 v2
c 2
v2 a
c 2 d
m 0 ac2
c 2 11 v2
c 2 3 2 2
m 0a
, as required.
11 v2
c
2 3 2 2
Section 3.2, pp. 135–138
1. a. The algorithm can be used; the
function is continuous.
b. The algorithm cannot be used; the
function is discontinuous at x 2.
c. The algorithm cannot be used; the
function is discontinuous at x 2.
d. The algorithm can be used; the
function is continuous on the given
domain.
2. a. max: 8, min: 12
b. max: 30, min: 5
c. max: 100, min: 100
d. max: 30, min: 20
3
2
3. a. max is 3 at x 0,
min is 1 at x 2
3
y
2
1
x
–2 –1 0 1 2 3
–1
–2
b. max is 4 at x 0,
min is 0 at x 2
y
6
4
2
x
–4 –2 0 2 4 6
–2
c. min is 4 at x 1, 2,
max is 0 at x 0, 3
y
2
x
–2 0 2 4
–2
–4
–4
–6
d. max is 0 at x 0,
min is 20 at x 2
y
8
x
–4 –2 0 2
–8
–16
–24
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