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2
C
1
area =
(0.0020 s) (140 + 200) m/s = 0.34 m/s.
2
From 6 to 7 ms, region
D
has the shape of a triangle with area
2
D
1
(0.0010 s) (200 m/s )= 0.10 m/s.
2
Substituting these values into Eq. 226, with
v
0
=0 then gives
0 12 m/s 0.26 m/s 0.34 m/s 0.10 m/s
0.82 m/s.
unhelmeted
v.
=+++=
Carrying out similar calculations for the helmeted head, we have the following results:
From 0 to 3 ms, region A has the shape of a triangle with area
2
A
1
(0.0030 s) (40 m/s ) = 0.060 m/s.
2
From 3 ms to 4 ms, region
B
has the shape of a rectangle with area
2
B
area = (0.0010 s) (40 m/s )= 0.040 m/s.
From 4 to 6 ms, region
C
has the shape of a trapezoid with area
2
C
1
(0.0020 s) (40 + 80) m/s = 0.12 m/s.
2
From 6 to 7 ms, region
D
has the shape of a triangle with area
2
D
1
(0.0010 s) (80 m/s )= 0.040 m/s.
2
69. The problem is solved using Eq. 226:
10
01
area between the acceleration curve
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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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