This preview shows page 1. Sign up to view the full content.
6. We apply the equation
2
1
00
2
()
x
tx
v
ta
t
=+
+
, found in Table 21. Since at
t
= 0 s,
x
0
= 0
and
0
12 m/s
v
=
, the equation becomes (in unit of meters)
2
1
2
() 12
x
tt
a
t
.
With
10 m
x
=
when
1.0 s
t
=
, the acceleration is found to be
2
4.0 m/s
a
=−
. The fact
that
0
a
<
implies that the bead is decelerating. Thus, the position is described by
2
() 12 2
.0
x
ttt
. Differentiating
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Acceleration

Click to edit the document details