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Unformatted text preview: 21 Chapter 2 1. We use Eq. 2-2 and Eq. 2-3. During a time t c when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with ∆ x = v t c . (a) During the first part of the motion, the displacement is ∆ x 1 = 40 km and the time interval is t 1 40 133 = = ( . km) (30 km / h) h. During the second part the displacement is ∆ x 2 = 40 km and the time interval is t 2 40 0 67 = = ( . km) (60 km / h) h. Both displacements are in the same direction, so the total displacement is ∆ x = ∆ x 1 + ∆ x 2 = 40 km + 40 km = 80 km. The total time for the trip is t = t 1 + t 2 = 2.00 h. Consequently, the average velocity is v avg km) (2.0 h) km / h. = = (80 40 (b) In this example, the numerical result for the average speed is the same as the average velocity 40 km/h. (c) As shown below, the graph consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to ( t 1 , x 1 ) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting ( t 1 , x 1 ) to ( t, x ) = (2.00 h, 80 km). From the graphical point of view, the slope of the dashed line drawn from the origin to ( t, x ) represents the average velocity. CHAPTER 2 22 2. Average speed, as opposed to average velocity, relates to the total distance, as opposed to the net displacement. The distance D up the hill is, of course, the same as the distance down the hill, and since the speed is constant (during each stage of the motion) we have speed = D / t . Thus, the average speed is D D t t D D v D v up down up down up down + + = + 2 which, after canceling D and plugging in v up = 40 km/h and v down = 60 km/h, yields 48 km/h for the average speed. 3. The speed (assumed constant) is v = (90 km/h)(1000 m/km) ∕ (3600 s/h) = 25 m/s. Thus, in 0.50 s, the car travels (0.50 s)(25 m/s) ≈ 13 m. 4. Huber’s speed is v 0 = (200 m)/(6.509 s)=30.72 m/s = 110.6 km/h, where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat Huber by 19.0 km/h, his speed is v 1 =(110.6 km/h + 19.0 km/h)=129.6 km/h, or 36 m/s (1 km/h = 0.2778 m/s). Thus, the time through a distance of 200 m for Whittingham is 1 200 m 5.554 s. 36 m/s x t v ∆ ∆ = = = 5. Using x = 3 t – 4 t 2 + t 3 with SI units understood is efficient (and is the approach we will use), but if we wished to make the units explicit we would write x = (3 m/s) t – (4 m/s 2 ) t 2 + (1 m/s 3 ) t 3 . We will quote our answers to one or two significant figures, and not try to follow the significant figure rules rigorously. (a) Plugging in t = 1 s yields x = 3 – 4 + 1 = 0. (b) With t = 2 s we get x = 3(2) – 4(2) 2 +(2) 3 = –2 m. (c) With t = 3 s we have x = 0 m....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
- Summer '10