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31. We assume the periods of acceleration (duration
t
1
) and deceleration (duration
t
2
) are
periods of constant
a
so that Table 21 can be used. Taking the direction of motion to be
+
x
then
a
1
= +1.22 m/s
2
and
a
2
= –1.22 m/s
2
. We use SI units so the velocity at
t
=
t
1
is
v
=
305/60 = 5.08 m/s.
(a) We denote
Δ
x
as the distance moved during
t
1
, and use Eq. 216:
2
22
01
2
(5.08 m/s)
2
2(1.22 m/s )
vv
a
x
x
=+
Δ
¡
Δ=
10.59 m 10.6 m.
=≈
(b) Using Eq. 211, we have
0
1
2
1
5.08 m/s
4.17 s.
1.22 m/s
t
a
−
==
=
The deceleration time
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics, Acceleration

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