Ch 2 Physics for Scientists and Engineers

# Ch 2 Physics for Scientists and Engineers - 2.1 Solve Model...

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2.1. Model: The car is represented by the particle model as a dot. Solve: (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 90 (b)

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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.

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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.
2.5. Model: We will consider the car to be a particle that occupies a single point in space. Visualize: Solve: Since the velocity is constant, we have xxv t x fi =+∆ . Using the above values, we get x 1 01 0 4 5 =+ = m m/s s 450 m () ( ) Assess: 10 m/s 22 mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is reasonable and expected.

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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: v ss tt s fi = (a) For the interval from the house to the lamppost: v 1 907 905 = =− 200 yd 600 yd 200 yd min :: / For the interval from the lamppost to the tree: v 2 910 907 = =+ 1200 yd 200 yd 333 yd min / (b) For the average velocity for the entire run: v avg 1200 yd 600 yd 120 yd min = 910 905 /
2.7. Model: Cars will be treated by the particle model. Visualize: Solve: Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: v x t xx tt v == ⇒=+ 10 Using the known values identified in the pictorial representation, we find: v v Alan 1 Alan 0 Alan 1 Alan 0 Beth 1 Beth 0 Beth 1 Beth 0 8:00 AM 400 mile 50 miles/hour AM 8 hr 4:00 PM 9:00 AM 400 mile 60 miles/hour 9:00 AM 6.67 hr 3:40 PM =+ = = 800 : (a) Beth arrives first. (b) Beth has to wait t Alan 1 t Beth 1 = 20 minutes for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem.

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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 v s t == = 100 m 50 m 20 s 2.5 m/s The slope at t = 25 s is v = = 100 m 100 m 10 s 0 m/s The slope at t = 35 s is v = =− 0 m 100 m 10 s 10 m/s
2.9.

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Ch 2 Physics for Scientists and Engineers - 2.1 Solve Model...

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