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# solution_pdf5 - garcia (jg46766) Problem Set 2 lacosse...

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Unformatted text preview: garcia (jg46766) Problem Set 2 lacosse (S210-7) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Do not wait until the last minute! 001 10.0 points When an objects acceleration is zero at some instant t in time, its velocity is 1. positive at t . 2. zero at t . 3. not changing at any time. 4. not changing at t . correct 5. negative at t . Explanation: By definition, the acceleration a = v t Since the acceleration is zero, v t = 0 v = 0 . 002 10.0 points The average speed of an orbiting space shuttle is 19800 mi / h. The shuttle is orbiting about 128 mi above the Earths surface. Assume the Earths radius is 3963 mi. How long does it take to circle the earth? Correct answer: 1 . 29821 h. Explanation: Let : v = 19800 mi / h , h = 128 mi , and r = 3963 mi . The distance traveled by the space shuttle in one orbit is 2 r = 2 ( R e + h ) , so the required time is t = 2 ( R e + h ) v s = 2 (3963 mi + 128 mi) 19800 mi / h = 1 . 29821 h . 003 10.0 points A 500-kilogram sports car accelerates uni- formly from rest, reaching a speed of 30 me- ters per second in 6 seconds. What distance has the car traveled during the 6 seconds? 1. 30 m 2. 15 m 3. 180 m 4. 60 m 5. 90 m correct Explanation: Let : v f = 30 m / s and t = 6 s . Since v = 0 and s = 0, v f = v + a t = a t a = v f t . s = s t + v t + 1 2 a t 2 = 1 2 a t 2 = 1 2 parenleftBig v f t parenrightBig t 2 = 1 2 v f t = 1 2 (30 m / s) (6 s) = 90 m . 004 (part 1 of 2) 10.0 points An acceleration (in m/s 2 ) has the time de- pendence shown on the graph below. The garcia (jg46766) Problem Set 2 lacosse (S210-7) 2 particle starts from rest ( v = 0 m/s) at the origin ( x = 0 m). 1 2 3 4 5- 3- 2- 1 1 2 3 a (m/s 2 ) t (s) Acceleration vs Time Find the velocity at t = 5 s. 1. v t =5 = 6 m / s 2. v t =5 =- 8 m / s correct 3. v t =5 = 3 m / s 4. v t =5 = 9 m / s 5. v t =5 = 2 m / s 6. v t =5 =- 1 m / s 7. v t =5 = 4 m / s 8. v t =5 =- 2 m / s 9. v t =5 =- 3 m / s 10. v t =5 =- 6 m / s Explanation: Let : t = 0 s , t 4 = 4 s , t 5 = 5 s , a 4 =- 2 m / s 2 , and a 4 5 = 0 m / s 2 . If the particle has a negative acceleration, its velocity will decrease from zero at the be- ginning to a negative value. When the accel- eration is zero, the velocity of the particle will be constant. Thus, the velocity of the particle is v 4 = v + a 4 ( t 4- t ) = 0 m / s + (- 2 m / s 2 ) (4 s- 0 s) =- 8 m / s , so v 5 = v 4 + a 4 5 ( t 5- t 4 ) =- 8 m / s + (0 m / s 2 ) (5 s- 4 s) =- 8 m / s . 005 (part 2 of 2) 10.0 points Find the position at t = 5 s. 1. x t =5 =- 7 m 2. x t =5 =- 24 m correct 3. x t =5 = 16 m 4. x t =5 =- 19 m 5. x t =5 =- 3 . 5 m 6. x t =5 =- 16 m 7. x t =5 =- 23 m 8. x t =5 = 23 . 5 m 9. x t =5 = 24 m 10. x t =5 = 1 m Explanation: The particle will move in a negative di- rection from rest at a negative acceleration....
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## solution_pdf5 - garcia (jg46766) Problem Set 2 lacosse...

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