fisica1-youn-e-freedman-exercicios-resolvidos

2.34: After the initial acceleration, the train has traveled
1 2
2
(1.60
2
m s
)(14.0 s)
= 156.8 m
(from Eq. (2.12), with x 0 = 0, v 0x = 0), and has attained a speed of
2
(1.60 m s )(14.0 s) = 22.4 m s.
During the 70-second period when the train moves with constant speed, the train travels
22 .4 m s 70 s = 1568 The distance traveled during deceleration is given by Eq.
( )( ) m.
x
=
x
(2.13), with v 0,
v 0
= 22.4 m s and
a
x
= −3.50 m
2
s
, so the train moves a distance
2
−(22.4
m / s)
x − x = = 71.68 m. The total distance covered in then 156.8 m + 1568 m + 71.7 m
0 2
2( −3.50
m/s )
= 1.8 km.
In terms of the initial acceleration a 1 , the initial acceleration time t 1 , the time t 2 during
which the train moves at constant speed and the magnitude a 2 of the final acceleration,
the total distance x T is given by
x
1
= a1t
2
2
1
which yields the same result.
T
+ ( a t ) t
1 1
2
+
2
1 ( a1t1
)
2 | a |
2
⎛
= ⎜
⎝
a1t
2
1
⎞⎛
⎟
⎜t
⎠⎝
1
+ 2t
2
a1t1
,
|
2
| ⎟ ⎞
+
a ⎠