If
we
know thexacce1eration
ax
as a function
of
time and
we
know the initial velocity
VOx,
we
can use Eq. (2.17)
to
find the xvelocity
Vx
at
any time; in other words,
we
can find
Vx
as a function
of
time. Once
we
know this function, and given the initial position
xo,
we
can
use
Eq. (2.18)
to
find the position
x
at
any time.
Motion
with
changing
acceleration
Sally
is
driving along a straight highway
in
her
classic 1965 Mus
tang.
At
tinte
t
= 0,
when
Sally
is
moving
at
10
m/s
in
the positive
xdirection, she passes a signpost
at
x
=
50
m.
Her
xacceleration
is
a function
of
tinte:
ax
=
2.0m/sz

(O.lOm/s')t
(a)
Find
her
xvelocity and position
as
functions
of
tinte.
(b)
When
is
her
xvelocity greatest? (c)
What
is
the
maximwn
xvelocity?
(d) Where
is
the
car
when
it
reaches
the
maximwn
xvelocity?
IDENTIFY:
The
xacceleration
is
a function
of
tinte,
so
we
cannot
use
the constantacceleration formulas
of
Section 2.4.
SET
UP:
We
use
Eqs. (2.17)
and
(2.18)
to
find
the
xvelocity and
position as functions
of
time. Once
we
have those functions,
we'll
be
able to answer a variety
of
questions about
the
motion.
EXECUTE:
(a)
At
t
= 0, Sally's position
is
Xo
=
50
m
and
her
xvelocity
is
v",
= 10
m/s.
Since
we
are given
the
xacceleration
ax
as a function
of
time,
we
firstuseEq.
(2.17) to find the xvelocity
Vx
as
a function
of
tinte
t.
The
integral
of
t"
is
Jt"
dt
= "
!
1
t"+1
for
n
*
1,so
Vx
=
lOm/s
+
I'[2.0m/s'

(O.lOm/s')tldt
o
1
=
lOm/s
+
(2.0m/s')t

2(0.lOm/s')t'
Then
we
use
Eq. (2.18)
to
find
x
as
a function
of
t:
x
=
50m
+
fJ
lOm/s
+
(2.0m/s')t

~(O.lOm/s')t'l
dt
1
I
=
50m
+
(lOm/s)t
+
2(2.0m/s')t'

6(0.lOm/s')t'
Figure 2.29 shows graphs
of
a
..
v
..
and
x
as functions
of
time.
Note that
for
any
tinte
t,
the
slope
of
the
vxt
graph equals
the
value
of
ax
and
the
slope
of
the
xt
graph equals
the
value
of
V
x
.
(b)
The
maximum value
ofv
x
occurs
when
the
xvelocity stops
increasing
and
begins to decrease.
At
this instant,
du,/dt
=
ax
=
O.
Setting
the
expression
for
ax
equal
to
zero,
we
obtsin
0=
2.0m/s'

(O.lOm/s')t
2.0m/s'
t=
=20s
O.lOm/s'
(c) We find
the
maximum xvelocity
by
substituting
t
=
20
s
(when xvelucity
is
maximwn) into
the
equation for
Vx
from
part
(a):
1
v""",.x
=
lOm/s
+
(2.0m/s')(20s)

2(0.1Om/s')(20s)'
=
30m/s
2.29
The
position, velocity, and acceleration
of
the car
in
Example 2.9 as functions
of
time. Can you show that
if
this
motion continues, the car
will
stop at
t
=
44.5
s?
ax
(mfs2)
2.0
L
......
xaccelemtion
is
~
~~ti~~
before
t
=
20
s.
1.0
~
:::O+l.5..JI0IL52O'~'S..J30
t(s)
1.0
xacce1emtion
is
....
....
............
negative
after
t
=
20
s.
t=208.
0
5
10
15
t(s)
20
25
30
x(m)
800
600
400
25
30
t(8)
2.6
*Velocity
and
Position
by
Integration
59
(d) The maximum value
of
Vx
occurs
at
time
t
=
20
s. To obtain
the position
of
the car at that time,
we
substitnte
t
=
20
s into the
expression for
x
from part (a):
I
x
=
50m
+
(1Om/s)(20
s)
+
2:(2.0m/
S
2)(20
S)2

~(O.IO
m/
s
3)(20
S)3
=
517m
EVALUATE:
Figure 2.29 helps us interpret
our
results.
The
top
graph
in
this figure shows that
ax
is
positive between
t
=
0 and
t
=
20
s and negative after that.
It
is
zero at
t
=
20
s, the time at
which
Vx
is maximum (the high point in the middle graph). The car
speeds up until
t
=
20
s (because
Vx
and
ax
have the same sign)
and slows down
after
t
=
20
s (because
Vx
and
ax
have opposite
signs).
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 Acceleration