(B)
Calculate the average velocity during these two time in
tervals.
Solution
In
the
fi
rst
time
interval,
t
t
f
t
i
t
B
t
A
1 s. Therefore, using Equation 2.2 and the dis
placement calculated in (a), we
fi
nd that
2 m/s
v
x
(
A
:
B
)
x
A
:
B
t
2 m
1 s
8 m
[
4(3)
2(3)
2
]
[
4(1)
2(1)
2
]
x
B
:
D
x
f
x
i
x
D
x
B
2 m
[
4(1)
2(1)
2
]
[
4(0)
2(0)
2
]
x
A
:
B
x
f
x
i
x
B
x
A
30
CHAPTER 2
•
Motion in One Dimension
Conceptual Example 2.2
The Velocity of Different Objects
Example 2.3
Average and Instantaneous Velocity
Consider the following onedimensional motions:
(A)
A ball
thrown directly upward rises to a highest point and falls
back into the thrower
’
s hand.
(B)
A race car starts from rest
and speeds up to 100 m/s.
(C)
A spacecraft drifts through
space at constant velocity. Are there any points in the mo
tion of these objects at which the instantaneous velocity has
the same value as the average velocity over the entire mo
tion? If so, identify the point(s).
Solution
(A) The average velocity for the thrown ball is
zero because the ball returns to the starting point; thus its
displacement is zero. (Remember that average velocity is
de
fi
ned as
x
/
t
.) There is one point at which the instanta
neous velocity is zero
—
at the top of the motion.
(B) The car
’
s average velocity cannot be evaluated unam
biguously with the information given, but it must be some
value between 0 and 100 m/s. Because the car will have
every instantaneous velocity between 0 and 100 m/s at
some time during the interval, there must be some instant
at which the instantaneous velocity is equal to the average
velocity.
(C) Because the spacecraft
’
s instantaneous velocity is con
stant, its instantaneous velocity at
any
time and its average
velocity over
any
time interval are the same.
10
8
6
4
2
0
–
2
–
4
0
1
2
3
4
t
(s)
x
(m)
Slope = 4 m/s
Slope =
–
2 m/s
Figure 2.4
(Example 2.3) Position
–
time graph for a particle
having an
x
coordinate that varies in time according to the
expression
x
4
t
2
t
2
.
In the second time interval,
t
2 s; therefore,
These values are the same as the slopes of the lines joining
these points in Figure 2.4.
(C)
Find the instantaneous velocity of the particle at
t
2.5 s.
Solution
We can guess that this instantaneous velocity must
be of the same order of magnitude as our previous results,
that is, a few meters per second. By measuring the slope of
the green line at
t
2.5 s in Figure 2.4, we
fi
nd that
6 m/s
v
x
4 m/s
v
x
(
B
:
D
)
x
B
:
D
t
8 m
2 s
3
Simply to make it easier to read, we write the expression as
x
4
t
2
t
2
rather than as
x
(
4.00 m/s)
t
(2.00 m/s
2
)
t
2.00
.
When an equation summarizes measurements, consider its coef
fi

cients to have as many signi
fi
cant digits as other data quoted in a
problem. Consider its coef
fi
cients to have the units required for di
mensional consistency. When we start our clocks at
t
0, we usually
do not mean to limit the precision to a single digit. Consider any
zero value in this book to have as many signi
fi
cant
fi
gures as you
need.
v
SECTION 2.3
•
Acceleration
31
(a) A car, modeled as a particle, moving along the
x
has velocity
v
xi
at
t
t
i
and velocity
v
xf
at
t
t
f
.
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 Spring '11
 WANG
 Acceleration, Velocity