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Motion Along a Straight Line
211
EVALUATE:
When
a
!
and
v
!
are in the same direction, the speed increases (
0
t
=
to
8 s
t
=
). When
a
!
and
v
!
are in
opposite directions, the speed decreases (
11 s
t
=
to
15 s
t
=
). When
0
a
=
the speed is constant
8 s
t
=
to
11 s
t
=
.
Figure 2.30ab
2.31.
(a) IDENTIFY
and
SET UP:
The acceleration
x
a
at time
t
is the slope of the tangent to the
x
v
versus
t
curve at
time
t
.
EXECUTE:
At
3 s,
t
=
the
x
v
versus
t
curve is a horizontal straight line, with zero slope. Thus
0.
x
a
=
At
7 s,
t
=
the
x
v
versus
t
curve is a straightline segment with slope
2
45 m/s
20 m/s
6.3 m/s .
9 s
5 s
−
=
−
Thus
2
6.3 m/s .
x
a
=
At
11 s
t
=
the curve is again a straightline segment, now with slope
2
04
5
m
/
s
11.2 m/s .
13 s 9 s
−−
=−
−
Thus
2
11.2 m/s .
x
a
EVALUATE:
0
x
a
=
when
x
v
is constant,
0
x
a
>
when
x
v
is positive and the speed is increasing, and
0
x
a
<
when
x
v
is positive and the speed is decreasing.
(b)
IDENTIFY:
Calculate the displacement during the specified time interval.
SET UP:
We can use the constant acceleration equations only for time intervals during which the acceleration is
constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration
equations for each segment. For the time interval
0
t
=
to
5 s
t
=
the acceleration is constant and equal to zero.
For the time interval
5 s
t
=
to
9 s
t
=
the acceleration is constant and equal to
2
6.25 m/s . For the interval
9 s
t
=
to
13 s
t
=
the acceleration is constant and equal to
2
11.2 m/s .
−
EXECUTE:
During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas
can be used.
0
20 m/s
x
v
=
0
x
a
=
5 s
t
=
0
?
xx
−=
00
x
x
xv
t
(0
x
a
=
so no
2
1
2
x
at
term)
0
(20 m/s)(5 s) 100 m;
=
this is the distance the officer travels in the first 5 seconds.
During the interval
5 s
t
=
to 9 s the acceleration is again constant. The constant acceleration formulas can be
applied to this 4 second interval. It is convenient to restart our clock so the interval starts at time
0
t
=
and ends at
time
5 s.
t
=
(Note that the acceleration is
not
constant over the entire
0
t
=
to
9 s
t
=
interval.)
0
20 m/s
x
v
=
2
6.25 m/s
x
a
=
4 s
t
=
0
100 m
x
=
0
?
−
=
2
1
2
x
x
x
t a
t
−= +
22
1
0
2
(20 m/s)(4 s)
(6.25 m/s )(4 s)
80 m 50 m 130 m.
+
=
+
=
Thus
0
130 m 100 m 130 m
230 m.
−+
=
+
=
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View Full Document212
Chapter 2
At
9 s
t
=
the officer is at
230 m,
x
=
so she has traveled 230 m in the first 9 seconds.
During the interval
9 s
t
=
to
13 s
t
=
the acceleration is again constant. The constant acceleration formulas can be
applied for this 4 second interval but
not
for the whole
0
t
=
to
13 s
t
=
interval. To use the equations restart our
clock so this interval begins at time
0
t
=
and ends at time
4 s.
t
=
0
45 m/s
x
v
=
(at the start of this time interval)
2
11.2 m/s
x
a
=−
4 s
t
=
0
230 m
x
=
0
?
xx
−
=
2
1
00
2
x
x
x
xv
t a
t
−= +
22
1
0
2
(45 m/s)(4 s)
( 11.2 m/s )(4 s)
180 m 89.6 m
90.4 m.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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