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31_Physics ProblemsTechnical Physics

31_Physics ProblemsTechnical Physics - 32*P2.26 Motion in...

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32 Motion in One Dimension *P2.26 The time for the Ford to slow down we find from x x v v t t x v v f i xi xf xi xf = + + = + = + = 1 2 2 2 250 71 5 0 6 99 d i a f m m s s . . . Its time to speed up is similarly t = ( ) + = 2 350 0 71 5 9 79 m m s s . . . The whole time it is moving at less than maximum speed is 6 99 5 00 9 79 21 8 . . . . s s s s + + = . The Mercedes travels x x v v t f i xi xf = + + = + + = 1 2 0 1 2 71 5 71 5 21 8 1 558 d i a fb ga f . . . m s s m while the Ford travels 250 350 600 + = m m, to fall behind by 1 558 600 958 m m m = . P2.27 (a) v i = 100 m s, a = 5 00 . m s 2 , v v at f i = + so 0 100 5 = t , v v a x x f i f i 2 2 2 = + c h so 0 100 2 5 00 0 2 =( ) ( ) . x f c h . Thus x f = 1 000 m and t = 20 0 . s . (b) At this acceleration the plane would overshoot the runway: No . P2.28 (a) Take t i = 0 at the bottom of the hill where x i = 0 , v i = 30 0 . m s, a = 2 00 . m s 2 . Use these values in the general equation x x v t at f i i = + + 1 2 2 to find x t t f = + + 0 30 0 1 2 2 00 2 . . m s m s
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