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oldfinal 02 JT10-solutions

# oldfinal 02 JT10-solutions - Bodet(ngb299 oldnal 02 JT10...

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Bodet (ngb299) – oldfinal 02 JT10 – turner – (57340) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A train pulls away from a station with a con- stant acceleration of 0 . 6 m / s 2 . A passenger arrives at a position on the track 6 s after the end of the train left that position. What is the slowest constant speed at which she can run and catch the train? Correct answer: 7 . 2 m / s. Explanation: Let : x 0 = 0 m , end of train position v 0 t = 0 m / s , a t = 0 . 6 m / s 2 , and Δ t = 6 s . After time t the train has traveled a dis- tance of x t = x 0 + v 0 t t + 1 2 a t t 2 = 1 2 a t t 2 and the passenger a distance of x p = x 0 + v p ( t - Δ t ) = v p ( t - Δ t ) . When the passenger overtakes the train, x t = x p v p ( t - Δ t ) = 1 2 a t t 2 v p = a t t 2 2 ( t - Δ t ) . To minimize v p , d dt a t t 2 2 ( t - Δ t ) = 0 4 ( t - Δ t ) a t t - 2 a t t 2 4 ( t - Δ t ) 2 = 0 2 ( t - Δ t ) - t = 0 t - 2 Δ t = 0 t = 2 Δ t = 2 (6 s) = 12 s , so v p = a t t 2 2 ( t - Δ t ) = a t (2 Δ t ) 2 2 Δ t = 2 a t Δ t = 2 (0 . 6 m / s 2 ) (6 s) = 7 . 2 m / s . 0 4 8 12 16 0 5 10 15 20 25 30 35 40 45 50 x (m) t (s) Δ t Train Passenger The train’s velocity at this time is v t = a t t = 2 a t Δ t = 2 (0 . 6 m / s 2 ) (6 s) = 7 . 2 m / s . The dashed line (representing the passen- ger’s minimum constant velocity) is tangent to the train’s acceleration curve, illustrating the fact that the velocities are equal at the passenger’s minimum speed. 002 10.0 points Two bodies are falling with negligible air resis- tance, side by side, above a horizontal plane. If one of the bodies is given an additional horizontal acceleration during its descent, it 1. has the vertical component of its acceler- ation altered. 2. has the vertical component of its velocity altered. 3. strikes the plane at the same time as the other body. correct

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Bodet (ngb299) – oldfinal 02 JT10 – turner – (57340) 2 4. follows a straight line path along the re- sultant acceleration vector. 5. follows a hyperbolic path. Explanation: Because a horizontal acceleration does not change the vertical velocity of an object, the two bodies strike the plane at the same time. 003 10.0 points A man stands in an elevator in the university’s administration building and is accelerating upwards. (During peak hours, this does not happen very often.) Elevator Cable Choose the correct free body diagram for the man, where F i,j is the force on the object i , from the object j . 1. F elevator, cable F man, elevator 2. F F man, floor man, earth correct 3. F F elevator, cable man, earth 4. F man, floor F elevator, cable F man, earth 5. F man, acceleration 6. F man, cable Explanation: Only the forces acting directly on the man are to be in the free body diagram. Therefore, the force from the cable should be omitted, while those from gravity and from the floor’s normal force should be included. 004 10.0 points The horizontal surface on which the objects slide is frictionless.
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