problem02_95 solution

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

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2.95: For convenience, let the student's (constant) speed be v 0 and the bus's initial position be x 0 . Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial velocity of the bus is taken to be zero. The positions of the student x 1 and the bus x 2 as functions of time are then , 0 1 t v x = . ) 2 1 ( 2 0 2 at x x + = a) Setting 2 1 x x = and solving for the times t gives ( 29 0 2 0 0 2 1 ax v v a t - ± = ( 29 ) m 0 . 40 )( s m 170 . 0 ( 2 ) s m 0 . 5 ( ) s m 0 . 5 ( ) s m 170 . 0 ( 1 2 2 2 - ± = s. 3 49 s, 55 . 9 . = The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later time). During this time, the student has run a distance m. 8 47 s) 55 . 9 )( s m 5 ( 0 . t v = = b) The speed of the bus is . s m 1.62 s) 55 . 9 )( s m 170 . 0 ( 2 = c) The results can be verified by noting that the x lines for the student and the bus intersect at two points: d) At the later time, the student has passed the bus, maintaining her constant speed,
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Unformatted text preview: but the accelerating bus then catches up to her. At this later time the bus's velocity is ( 29 ( 29 . s m 38 . 8 s 3 . 49 s m 170 . 2 = e) No; , 2 2 ax v < and the roots of the quadratic are imaginary. When the student runs at , s m 5 . 3 , the two lines do not intersect: f) For the student to catch the bus, . 2 2 ax v and so the minimum speed is ( 29 ( 29 s. m 688 . 3 s m 40 s m 170 . 2 2 = She would be running for a time s, 7 . 21 2 m/s 0.170 s m 69 . 3 = and cover a distance ( 29 ( 29 m. . 80 s 7 . 21 s m 688 . 3 = However, when the student runs at s, m 688 . 3 the lines intersect at one point ( x = 80 m)....
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problem02_95 solution - but the accelerating bus then...

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