HW 1 solutions

HW 1 solutions - Chapter 2 Problems 14 18 22 23 30 36 51 52...

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Chapter 2 Problems: 14, 18, 22, 23, 30, 36, 51, 52, 59, 61, 78 14 . (a) The average speed is v = 4800 km= 11 h = 436 : 4 km=h: (b) The plane travels at an average speed of v air = 960 km=hr; the total distance covered is s = 6700 km: The total time in the air is t air = s=v air = 6700 = 960 = 6 : 98 h: Hence the time on the ground is t g 4 h: 18 sec v avg = x t = 2 : 5 2 = 1 : 25 m=s: (b) After 4 sec v avg = x t = 0 4 = 0 m=s: (c) After 6 sec v avg = x t = 2 6 = : 333 m=s: (d) During the interval from t = 3 sec and t = 4 sec v avg = x t = 0 3 1 = 3 m=s: 23 . (a) The velocity for the rocket as a function of time is v = dy dt = d dt bt ct 2 ± = b 2 bt = (82 9 : 8 t ) m=s 2 : (b) The velocity is zero when t = 82 = 9 : 8 = 8 : 4 s: 30 . The average acceleration during the fall is a v= t = 11 = 1 : 12 = 9 : 82 m=s 2 : The average acceleration on contact with the ground is a v= t = 11 =: 131 = 84 : 0 m=s 2 1

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36 . Given that x = bt 3 ; where b = 1 : 5 m=s 3 ; (a) v at t = 2 : 5 sec ; v = dx dt = 3 bt 2 = 3 (1 : 5) (2 : 5) 2 = 28 : 125 m=s; (b) a at t = 2 : 5 sec ; a = dv dt = 6 bt = 6 (1 : 5) 2 : 5 = 22 : 5 m=s 2 ; (c) the average velocity during this time is h v i x= t = bt 3 =t = bt 2 = (1 : 5) (2 : 5) 2 = 9 : 375 m=s; (d) the average acceleration during this time is h a i v= t = 3 bt 2 =t = 3 bt = 4 : 5 2 : 5 = 11 : 25 m=s 2 : 51 . The acceleration of the airliner is a = ± v t = ± 220 10 3 3600 1 29 = ± 2 : 107 m=s 2 Hence the distance required to stop is x = v i t + 1 2 at 2 = 220 10 3 3600 29 ± 1 2 (2 : 107) 29 2 = 886 m: Since the acceleration is constant this distance can also be found from
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HW 1 solutions - Chapter 2 Problems 14 18 22 23 30 36 51 52...

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