Phys 141_HW Sol_Ch2_W08

Phys 141_HW Sol_Ch2_W08 - Cal Poly State University, SLO...

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Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 Homework Solutions Chapter 2 Assigned Problems: 2, 3, 4, 6, 8, 16, 18, 21, 24, 30, 32, 52, 70, 78, 82 2.2. Solve: Diagram Position Velocity Acceleration (a) (b) (c) Negative Negative Positive Positive Negative Negative Positive Negative Negative 2.3. Solve: A forgetful physics professor goes for a walk on a straight country road. Walking at a constant speed, he covers a distance of 300 m in 300 s. He then stops and watches the sunset for 100 s. Finding that it was getting dark, he walks faster back to his house covering the same distance in 200 s. 2.4. Solve: Forty miles into a car trip north from his home in El Dorado, an absent-minded English professor stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he was supposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one hour of driving at another rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back very slowly, confused and tired as he was, and reached El Dorado in two hours. 2.6. Model: We will consider Larry to be a particle. Visualize: Solve: Since Larry’s speed is constant, we can use the following equation to calculate the velocities: f i s f i s s v t t - = - (a) For the interval from the house to the lamppost: 1 200 yd 600 yd 200 yd/min 9:07 9:05 v - = = - - For the interval from the lamppost to the tree: 2 1200 yd 200 yd 333 yd/min 9:10 9:07 v - = = + -
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Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 (b) For the average velocity for the entire run: avg 1200 yd 600 yd 120 yd/min 9:10 9:05 v - = = + - 2.8. Model: The bicyclist is a particle. Visualize: Please refer to Figure Ex2.8. Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t = 10 s is 100 m 50 m 2.5 m/s 20 s s v t - = = = The slope at t = 25 s is 100 m 100 m 0 m/s 10 s v - = = The slope at t = 35 s is 0 m 100 m 10 m/s 10 s v - = = - 2.16. Model: Represent the jet plane as a particle Visualize: Solve: (a) Since we don’t know the time of acceleration, we will use 2 2 1 0 1 0 2 2 2 2 2 1 0 1 2 ( ) (400 m/s) (300 m/s) 8.75 m/s 2 2(4000 m) v v a x x v v a x = + - - - = = = (b) The acceleration of the jet is approximately equal to g , the acceleration due to gravity. 2.18.
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This note was uploaded on 04/02/2008 for the course PHYS 141 taught by Professor Staff during the Winter '06 term at Cal Poly.

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Phys 141_HW Sol_Ch2_W08 - Cal Poly State University, SLO...

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