41_PartUniversity Physics Solution

41_PartUniversity - Motion Along a Straight Line 2-11 EVALUATE When a and v are in the same direction the speed increases t = 0 to t = 8 s When a

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Motion Along a Straight Line 2-11 EVALUATE: When a ! and v ! are in the same direction, the speed increases ( 0 t = to 8 s t = ). When a ! and v ! are in opposite directions, the speed decreases ( 11 s t = to 15 s t = ). When 0 a = the speed is constant 8 s t = to 11 s t = . Figure 2.30a-b 2.31. (a) IDENTIFY and SET UP: The acceleration x a at time t is the slope of the tangent to the x v versus t curve at time t . EXECUTE: At 3 s, t = the x v versus t curve is a horizontal straight line, with zero slope. Thus 0. x a = At 7 s, t = the x v versus t curve is a straight-line segment with slope 2 45 m/s 20 m/s 6.3 m/s . 9 s 5 s = Thus 2 6.3 m/s . x a = At 11 s t = the curve is again a straight-line segment, now with slope 2 04 5 m / s 11.2 m/s . 13 s 9 s −− =− Thus 2 11.2 m/s . x a EVALUATE: 0 x a = when x v is constant, 0 x a > when x v is positive and the speed is increasing, and 0 x a < when x v is positive and the speed is decreasing. (b) IDENTIFY: Calculate the displacement during the specified time interval. SET UP: We can use the constant acceleration equations only for time intervals during which the acceleration is constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration equations for each segment. For the time interval 0 t = to 5 s t = the acceleration is constant and equal to zero. For the time interval 5 s t = to 9 s t = the acceleration is constant and equal to 2 6.25 m/s . For the interval 9 s t = to 13 s t = the acceleration is constant and equal to 2 11.2 m/s . EXECUTE: During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas can be used. 0 20 m/s x v = 0 x a = 5 s t = 0 ? xx −= 00 x x xv t (0 x a = so no 2 1 2 x at term) 0 (20 m/s)(5 s) 100 m; = this is the distance the officer travels in the first 5 seconds. During the interval 5 s t = to 9 s the acceleration is again constant. The constant acceleration formulas can be applied to this 4 second interval. It is convenient to restart our clock so the interval starts at time 0 t = and ends at time 5 s. t = (Note that the acceleration is not constant over the entire 0 t = to 9 s t = interval.) 0 20 m/s x v = 2 6.25 m/s x a = 4 s t = 0 100 m x = 0 ? = 2 1 2 x x x t a t −= + 22 1 0 2 (20 m/s)(4 s) (6.25 m/s )(4 s) 80 m 50 m 130 m. + = + = Thus 0 130 m 100 m 130 m 230 m. −+ = + =
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2-12 Chapter 2 At 9 s t = the officer is at 230 m, x = so she has traveled 230 m in the first 9 seconds. During the interval 9 s t = to 13 s t = the acceleration is again constant. The constant acceleration formulas can be applied for this 4 second interval but not for the whole 0 t = to 13 s t = interval. To use the equations restart our clock so this interval begins at time 0 t = and ends at time 4 s. t = 0 45 m/s x v = (at the start of this time interval) 2 11.2 m/s x a =− 4 s t = 0 230 m x = 0 ? xx = 2 1 00 2 x x x xv t a t −= + 22 1 0 2 (45 m/s)(4 s) ( 11.2 m/s )(4 s) 180 m 89.6 m 90.4 m.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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41_PartUniversity - Motion Along a Straight Line 2-11 EVALUATE When a and v are in the same direction the speed increases t = 0 to t = 8 s When a

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