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ch02-p069 - 69 The problem is solved using Eq 2-26 v1 v0 =...

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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 area = (0.0010 s) (200 m/s )= 0.10 m/s. 2 Substituting these values into Eq. 2-26, 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 area = (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 area = (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 area = (0.0010 s) (80 m/s )= 0.040 m/s. 2 69. The problem is solved using Eq. 2-26: 1 0 0 1 area between the acceleration curve
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