problem02_34 solution

University Physics with Modern Physics with Mastering Physics (11th Edition)

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2.34: After the initial acceleration, the train has traveled m 8 . 156 ) s 0 . 14 )( s m 60 . 1 ( 2 1 2 2 = (from Eq. (2.12), with x 0 = 0 , v 0 x = 0) , and has attained a speed of . s m 4 . 22 ) s 0 . 14 )( s m 60 . 1 ( 2 = During the 70-second period when the train moves with constant speed, the train travels ( 29 ( 29 m. 1568 s 70 s m 4 . 22 = The distance traveled during deceleration is given by Eq. (2.13), with s m 4 . 22 , 0 0 = = x x v v and 2 s m 50 . 3 - = x a , so the train moves a distance . m 68 . 71 ) m/s 3.50 2( ) s / m 4 . 22 ( 0 2 2 = = - - - x x The total distance covered in then 156.8 m + 1568 m + 71.7 m = 1.8 km. In terms of the initial acceleration
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Unformatted text preview: a 1 , the initial acceleration time t 1 , the time t 2 during which the train moves at constant speed and the magnitude a 2 of the final acceleration, the total distance x T is given by which yields the same result. , | | 2 2 | | ) ( 2 1 ) ( 2 1 2 1 1 2 1 1 1 2 2 1 1 2 1 1 2 1 1 T + + = + + = a t a t t t a a t a t t a t a x...
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