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98. We obtain the velocity by integration of the acceleration:
0
0
(6.1 1.2 ')
'
t
vv
td
t
−=
−
³
.
Lengths are in meters and times are in seconds. The student is encouraged to look at the
discussion in the textbook in §27 to better understand the manipulations here.
(a) The result of the above calculation is
2
0
=
+ 6.1
0.6 ,
t
t
−
where the problem states that
v
0
= 2.7 m/s. The maximum of this function is found by
knowing when its derivative (the acceleration) is zero (
a
= 0 when
t
= 6.1/1.2 = 5.1 s) and
plugging that value of
t
into the velocity equation above. Thus, we find
18 m/s
v
=
.
(b) We integrate again to find
x
as a function of
t
:
22
3
00
0
(
6.1
0.6
)
3.05
0.2
tt
x
xv
d
t
v
t
t
d
t
v
t
t
t
′′
′
′
=
+
−
=
+
−
³³
.
With
x
0
= 7.3 m, we obtain
x
= 83 m for
t
= 6.
This is the correct answer, but one has
the right to worry that it might not be; after all, the problem asks for the total distance
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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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