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46
Motion in One Dimension
*P2.60
Average speed of every point on the train as the first car passes Liz:
∆
∆
x
t
==
860
573
.
.
m
1.50 s
ms
.
The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway
through the next 1.10 s, the speed of the train is
782
.
.
m
1.10 s
=
. The time required for the speed
to change from 5.73 m/s to 7.82 m/s is
1
2
150
1
2
110
130
..
.
s
s
s
()
+
=
so the acceleration is:
a
v
t
x
x
−
=
∆
∆
160
.
.
s
2
.
P2.61
The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then
v
xi
=
104
.
mm d and
a
x
=
F
H
G
I
K
J
0132
.
mm d
w
. The increase in the length of the hair (i.e., displacement)
during a time of
t
500
350
w
d
i
s
∆
∆
xv
t a
t
x
xi
x
=+
⋅
1
2
1
2
0 132
35 0
5 00
2
.
.
.
mm d
d
mm d w
d
w
bg
af
a
f
or
∆
x
=
48 0
. mm.
P2.62
Let point 0 be at ground level and point 1 be at the end of the engine burn. Let
point 2 be the highest point the rocket reaches and point 3 be just before
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 Fall '11
 Staff
 Physics

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