Solution a = 2t-6 dv = a dt L v 0 dv = L t 0 (2t-6) dt v = t 2-6t ds = v dt L s 0 ds = L t 0 (t 2-6t) dt s = t 3 3-3t 2 When t = 6 s, v = 0 Ans. When t = 11 s, s = 80.7 m Ans. 12–1. Starting from rest, a particle moving in a straight line has an acceleration of a = (2t-6) m>s 2 , where t is in seconds. What is the particle's velocity when t = 6 s, and what is its position when t = 11 s? Ans: s = 80.7 m
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Solution a = 2t-6 dv = a dt L v 0 dv = L t 0 (2t-6) dt v = t 2-6t ds = v dt L s 0 ds = L t 0 (t 2-6t) dt s = t 3 3-3t 2 When t = 6 s, v = 0 Ans. When t = 11 s, s = 80.7 m Ans. 12–1. Starting from rest, a particle moving in a straight line has an acceleration of a = (2t-6) m>s 2 , where t is in seconds. What is the particle's velocity when t = 6 s, and what is its position when t = 11 s? Ans: s = 80.7 m
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Solution a = 2t-6 dv = a dt L v 0 dv = L t 0 (2t-6) dt v = t 2-6t ds = v dt L s 0 ds = L t 0 (t 2-6t) dt s = t 3 3-3t 2 When t = 6 s, v = 0 Ans. When t = 11 s, s = 80.7 m Ans. 12–1. Starting from rest, a particle moving in a straight line has an acceleration of a = (2t-6) m>s 2 , where t is in seconds. What is the particle's velocity when t = 6 s, and what is its position when t = 11 s? Ans: s = 80.7 m
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Ans. t = 26.7 s 15 = 0 + 0.5625t A : + B v = v 0 + a c t a c = 0.5625 m>s 2 15 2 = 0 2 + 2a c (200-0) A : + B v 2 = v 0 2 + 2a c (s-s 0) s = 200 m s 0 = 0 v = 15 m>s v 0 = 0 • 12–1. A car starts from rest and with constant acceleration achieves a velocity of when it travels a distance of 200 m. Determine the acceleration of the car and the time required. Ans. = 450 m = 0 + 0 + 1 2 (1) A30 2 B A : + B s = s 0 + v 0 t + 1 2 a c t 2 = 0 + 1(30) = 30 m>s A : + B v = v 0 + a c t t = 30 s s 0 = 0 v 0 = 0 a c = 1 m>s 2 12–2. A train starts from rest at a station and travels with a constant acceleration of. Determine the velocity of the train when and the distance traveled during this time. t = 30 s 1 m>s 2
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