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# 21 - fryer(vdf96 Kinematics Motion with constant...

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fryer (vdf96) – Kinematics Motion with constant acceleration – graves – (6) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A jet plane lands with a speed of 110 m / s and can decelerate uniformly at a maximum rate of 5.5 m/s 2 as it comes to rest. Can this plane land at an airport where the runway is 0.86 km long? Answer this by calculating. Correct answer: 1 . 1 km. Explanation: Let : v i = 110 m / s , v f = 0 m / s , and Δ a = - 5 . 5 m / s 2 . a avg = Δ v Δ t = v f - v i Δ t = - v i Δ t Δ t = - v i a avg Δ x = 1 2 ( v i + v f ) Δ t = 1 2 v i Δ t = - 1 2 v i 2 a avg = - 1 2 (110 m / s) 2 ( - 5 . 5 m / s 2 ) · 1 km 1000 m = 1 . 1 km . The plane cannot land at an airport with a runway that is 0 . 86 km long, because it needs 1 . 1 km at a minimum to stop. 002(part1of2)10.0points A car starts from rest and travels for 5.3 s with a uniform acceleration of +2.1 m/s 2 . The driver then applies the brakes, causing a uniform deceleration of 1.6 m/s 2 . If the brakes are applied for 2.2 s, how fast is the car going at the end of the braking period? Correct answer: 7 . 61 m / s. Explanation: Let : v i = 0 m / s , Δ t 1 = 5 . 3 s , a 1 = +2 . 1 m / s 2 , a 2 = - 1 . 6 m / s 2 , and Δ t 2 = 2 . 2 s . For the first time interval, v f 1 = v 1 + a 1 Δ t 1 = a 1 Δ t 1 = ( 2 . 1 m / s 2 ) (5 . 3 s) = 11 . 13 m / s . For the second time interval, v i 2 = v f 1 : v f 2 = v i 2 + a 2 Δ t 2 = 11 . 13 m / s + ( - 1 . 6 m / s 2 ) (2 . 2 s) = 7 . 61 m / s .

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21 - fryer(vdf96 Kinematics Motion with constant...

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