S08P2AChap2Sol

# S08P2AChap2Sol - Chapter 2 Problem Solutions (Easy) v = !r...

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Chapter 2 Problem Solutions (Easy) 2.3 v = ! r ! t = ( 26 + 385 = 1760 ) mi ( 2 + 756 = 3600 ) h = 11 . 9 mi h " 5 . 30 m/s , 2.14 (a) v = ! r = ! t = 4800 km = 11 h = 436 km/h. (b) The actual flight time is ! t = 6700 km = (960 km/h) = 6.98 h, so 11 h ! 6.98 h = 4.02 h is spent on the ground. 2.22 From the given function, x ( t ) = bt + ct 3 , and the definition of average velocity (Equation 2-1), v = ! x = ! t , one obtains: (a) v a = [ x (3 s) ! x (1 s)] = (3 s ! 1 s) = [(1.5 m/s)(3 s ! 1 s) + (0.64 m/s 2 )(9 s 3 ! 1 s 3 ] = 2 s = 9.82 m/s; (b) v b = [ x (2.5 s) ! x (1.5 s)] = (2.5 s ! 1.5 s) = 9.34 m/s; (c) v c = [ x (2.05 s) ! x (1.95 s)] = (2.05 s ! 1.95 s) = 9.18 m/s. (d) The instantaneous velocity (Equation 2-2b) is v ( t ) = dx = dt = b + 3 ct 2 . At t = 2.00 s, v (2 s) = 1.5 m/s + 3(0.64 m/s 3 )(2 s) 2 = 9.18 m/s. As the interval surrounding 2 s gets smaller, the average and instantaneous velocities agree better; the values in parts (c) and (d) differ by less than 0.02%. 2.30 The velocity (positive downward) changes from v 1 = 0 to v 2 = 11.0 m/s in 1.12 s while falling with average acceleration a = ! v= ! t = (11.0 m/s) = (1.12 s) = 9.82 m/s 2 . When stopping, the egg’s velocity changes from v 2 = 11.0 m/s to v 3 = 0 in 0.131 s, with average acceleration a = (0 ! 11.0 m/s = 0.131 s = ! 84.0 m/s 2 . (A negative sign means that the acceleration is upward; when the speed is decreasing, this is called a deceleration.) 2.36 (a) v ( t ) = dx = dt = 3 bt 2 , and v (2.5 s) = 3(1.5 m/s 3 )(2.5 s) 2 = 28.1 m/s . (b) a ( t ) = d dt = 6 bt , and a (2.5 s) = 6(1.5 m/s 3 ) (2.5 s) = 22.5 m/s 2 . (c) v = [ x (2.5 s) ! x (0)] = 2.5 s = (1.5 m/s 3 )(2.5 s) 2 = 9.38 m/s. (d) a = [ v (2.5 s) ! v (0)] = 2.5 s = 3 )(2.5 s) = 11.3 m/s 2 . 2.51 From Equation 2-9 with v = 0, we find x ! x 0 = 1 2 v 0 t = 1 2 (220 km/h)(29 h/3600) = 886 m (over half a mile).

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